(1)THE EQUAL VARIABLE METHOD VASILE CÎRTOAJE DEPARTMENT OFAUTOMATION ANDCOMPUTERS UNIVERSITY OFPLOIE ¸STI BUCURE ¸STI39, ROMANIA vcirtoaje@upg-ploiesti.ro Received 01 March, 2006

THE EQUAL VARIABLE METHOD

VASILE CÎRTOAJE

DEPARTMENT OFAUTOMATION ANDCOMPUTERS

UNIVERSITY OFPLOIE ¸STI

BUCURE ¸STI39, ROMANIA

vcirtoaje@upg-ploiesti.ro

Received 01 March, 2006; accepted 17 April, 2006 Communicated by P.S. Bullen

ABSTRACT. The Equal Variable Method (called alson−1Equal Variable Method on the Math- links Site - Inequalities Forum) can be used to prove some difficult symmetric inequalities in- volving either three power means or, more general, two power means and an expression of form f(x1) +f(x2) +· · ·+f(xn).

Key words and phrases: Symmetric inequalities, Power means, EV-Theorem.

2000 Mathematics Subject Classification. 26D10, 26D20.

1. STATEMENT OF RESULTS

In order to state and prove the Equal Variable Theorem (EV-Theorem) we require the follow- ing lemma and proposition.

Lemma 1.1. Let a, b, c be fixed non-negative real numbers, not all equal and at most one of them equal to zero, and letx≤y≤zbe non-negative real numbers such that

x+y+z =a+b+c, x^p+y^p+z^p =a^p+b^p+c^p,

wherep ∈ (−∞,0]∪(1,∞). Forp = 0, the second equation isxyz =abc > 0. Then, there exist two non-negative real numbersx1andx2 withx1 < x2 such thatx∈[x1, x2]. Moreover,

(1) ifx=x₁ andp≤0, then0< x < y=z;

(2) ifx=x1 andp > 1, then either0 = x < y≤z or0< x < y=z;

(3) ifx∈(x1, x2), thenx < y < z;

(4) ifx=x₂, thenx=y < z.

Proposition 1.2. Let a, b, cbe fixed non-negative real numbers, not all equal and at most one of them equal to zero, and let0≤x≤y≤z such that

x+y+z =a+b+c, x^p+y^p+z^p =a^p+b^p+c^p,

059-06

wherep∈(−∞,0]∪(1,∞). Forp= 0, the second equation is xyz =abc >0. Letf(u)be a differentiable function on(0,∞), such thatg(x) =f⁰

x^p−11

is strictly convex on(0,∞), and let

F₃(x, y, z) =f(x) +f(y) +f(z).

(1) If p ≤ 0, then F₃ is maximal only for 0 < x = y < z, and is minimal only for 0< x < y=z;

(2) Ifp >1and eitherf(u)is continuous atu= 0orlim

u→0f(u) = −∞, thenF3is maximal only for0< x=y < z, and is minimal only for eitherx= 0or0< x < y=z.

Theorem 1.3 (Equal Variable Theorem (EV-Theorem)). Let a1, a2, . . . , an (n ≥ 3) be fixed non-negative real numbers, and let0≤x₁ ≤x₂ ≤ · · · ≤x_nsuch that

x₁+x₂+· · ·+x_n=a₁+a₂+· · ·+a_n, x^p₁+x^p₂+· · ·+x^p_n=a^p₁+a^p₂+· · ·+a^p_n,

wherepis a real number,p6= 1. Forp= 0, the second equation isx₁x₂· · ·x_n =a₁a₂· · ·a_n >

0. Letf(u)be a differentiable function on(0,∞)such that g(x) = f⁰

x^p−11 is strictly convex on(0,∞), and let

F_n(x₁, x₂, . . . , x_n) = f(x₁) +f(x₂) +· · ·+f(x_n).

(1) If p≤ 0, thenF_nis maximal for 0< x₁ = x₂ = · · ·=xn−1 ≤x_n, and is minimal for 0< x₁ ≤x₂ =x₃ =· · ·=x_n;

(2) If p > 0 and either f(u) is continuous at u = 0 or lim

u→0f(u) = −∞, then F_n is maximal for 0 ≤ x₁ = x₂ = · · · = x_n−1 ≤ x_n, and is minimal for either x₁ = 0 or 0< x₁ ≤x₂ =x₃ =· · ·=x_n.

Remark 1.4. Let 0 < α < β. If the function f is differentiable on (α, β) and the function g(x) = f⁰

x^p−11

is strictly convex on (α^p−1, β^p−1) or (β^p−1, α^p−1), then the EV-Theorem holds true forx₁, x₂, . . . , x_n ∈(α, β).

By Theorem 1.3, we easily obtain some particular results, which are very useful in applica- tions.

Corollary 1.5. Let a₁, a₂, . . . , a_n (n ≥ 3) be fixed non-negative numbers, and let0 ≤ x₁ ≤ x₂ ≤ · · · ≤x_nsuch that

x₁+x₂+· · ·+x_n=a₁+a₂+· · ·+a_n, x²₁+x²₂+· · ·+x²_n=a²₁+a²₂+· · ·+a²_n.

Letf be a differentiable function on(0,∞)such thatg(x) =f⁰(x)is strictly convex on(0,∞).

Moreover, eitherf(x)is continuous atx= 0orlim

x→0f(x) =−∞. Then, F_n =f(x₁) +f(x₂) +· · ·+f(x_n)

is maximal for 0 ≤ x₁ = x₂ = · · · = xn−1 ≤ x_n, and is minimal for either x₁ = 0 or 0< x₁ ≤x₂ =x₃ =· · ·=x_n.

Corollary 1.6. Leta₁, a₂, . . . , a_n (n ≥ 3)be fixed positive numbers, and let0 < x₁ ≤ x₂ ≤

· · · ≤x_nsuch that

x₁+x₂+· · ·+x_n=a₁+a₂+· · ·+a_n, 1

x₁ + 1

x₂ +· · ·+ 1 x_n = 1

a₁ + 1

a₂ +· · ·+ 1 a_n. Let f be a differentiable function on (0,∞) such that g(x) = f⁰

√1 x

is strictly convex on (0,∞). Then,

F_n =f(x₁) +f(x₂) +· · ·+f(x_n)

is maximal for0< x₁ =x₂ =· · ·=x_n−1 ≤x_n, and is minimal for0< x₁ ≤x₂ =x₃ =· · ·= x_n.

Corollary 1.7. Leta₁, a₂, . . . , a_n (n ≥ 3)be fixed positive numbers, and let0 < x₁ ≤ x₂ ≤

· · · ≤xnsuch that

x1+x2+· · ·+xn=a1+a2+· · ·+an, x1x2· · ·xn=a1a1· · ·an. Letfbe a differentiable function on(0,∞)such thatg(x) =f^{0 1}_x

is strictly convex on(0,∞).

Then,

F_n =f(x₁) +f(x₂) +· · ·+f(x_n)

is maximal for0< x1 =x2 =· · ·=xn−1 ≤xn, and is minimal for0< x1 ≤x2 =x3 =· · ·= x_n.

Corollary 1.8. Let a₁, a₂, . . . , a_n (n ≥ 3) be fixed non-negative numbers, and let0 ≤ x₁ ≤ x₂ ≤ · · · ≤x_nsuch that

x₁+x₂+· · ·+x_n=a₁+a₂+· · ·+a_n, x^p₁+x^p₂+· · ·+x^p_n=a^p₁+a^p₂+· · ·+a^p_n, wherepis a real number,p6= 0andp6= 1.

(a) Forp < 0,P =x₁x₂· · ·x_n is minimal when0 < x₁ =x₂ =· · · = xn−1 ≤ x_n, and is maximal when0< x1 ≤x2 =x3 =· · ·=xn.

(b) Forp > 0,P =x1x2· · ·xn is maximal when0≤ x1 =x2 =· · · =xn−1 ≤ xn, and is minimal when eitherx₁ = 0or0< x₁ ≤x₂ =x₃ =· · ·=x_n.

Corollary 1.9. Let a₁, a₂, . . . , a_n(n ≥ 3)be fixed non-negative numbers, let0 ≤ x₁ ≤ x₂ ≤

· · · ≤x_nsuch that

x₁+x₂+· · ·+x_n=a₁+a₂+· · ·+a_n, x^p₁+x^p₂+· · ·+x^p_n=a^p₁+a^p₂+· · ·+a^p_n, and letE =x^q₁ +x^q₂+· · ·+x^q_n.

Case 1. p≤0 (p= 0yieldsx₁x₂· · ·x_n =a₁a₂· · ·a_n>0).

(a) Forq ∈ (p,0)∪(1,∞), E is maximal when0< x₁ = x₂ =· · · = xn−1 ≤ x_n, and is minimal when0< x₁ ≤x₂ =x₃ =· · ·=x_n.

(b) Forq∈ (−∞, p)∪(0,1),E is minimal when0< x₁ =x₂ =· · ·=xn−1 ≤x_n, and is maximal when0< x₁ ≤x₂ =x₃ =· · ·=x_n.

Case 2. 0< p <1.

(a) Forq ∈ (0, p)∪(1,∞), E is maximal when0≤ x₁ = x₂ =· · · = xn−1 ≤x_n, and is minimal when eitherx₁ = 0or0< x₁ ≤x₂ =x₃ =· · ·=x_n.

(b) Forq∈ (−∞,0)∪(p,1),E is minimal when0≤x₁ =x₂ =· · ·=xn−1 ≤x_n, and is maximal when eitherx₁ = 0or0< x₁ ≤x₂ =x₃ =· · ·=x_n.

Case 3. p > 1.

(a) Forq ∈ (0,1)∪(p,∞), E is maximal when0≤ x1 = x2 =· · · = xn−1 ≤xn, and is minimal when eitherx₁ = 0or0< x₁ ≤x₂ =x₃ =· · ·=x_n.

(b) Forq∈ (−∞,0)∪(1, p),E is minimal when0≤x₁ =x₂ =· · ·=x_n−1 ≤x_n, and is maximal when eitherx₁ = 0or0< x₁ ≤x₂ =x₃ =· · ·=x_n.

2. PROOFS

Proof of Lemma 1.1. Leta ≤b ≤c. Note that in the excluded casesa =b=canda =b = 0, there is a single triple(x, y, z)which verifies the conditions

x+y+z =a+b+c and x^p+y^p+z^p =a^p+b^p+c^p. Consider now three cases: p= 0,p <0andp > 1.

A. Casep= 0 (xyz =abc >0). LetS = ^a+b+c₃ andP = √³

abc, whereS > P >0by AM-GM Inequality. We have

x+y+z = 3S, xyz =P³,

and from0< x≤y≤z andx < z, it follows that0< x < P. Now let f =y+z−2√

yz.

It is clear thatf ≥0, with equality if and only ify=z. Writingf as a function ofx, f(x) = 3S−x−2P

rP x, we have

f⁰(x) = P x

rP

x −1>0,

and hence the functionf(x)is strictly increasing. Sincef(P) = 3(S−P) > 0, the equation f(x) = 0has a unique positive rootx₁,0< x₁ < P. Fromf(x)≥0, it follows thatx≥x₁. Sub-casex=x₁. Sincef(x) =f(x₁) = 0andf = 0impliesy =z, we have0< x < y=z.

Sub-casex > x₁. We havef(x)>0andy < z. Consider now thatyandz depend onx. From x+y(x) +z(x) = 3S andx·y(x)·z(x) = P³, we get1 +y⁰ +z⁰ = 0and ¹_x + ^y_y⁰ + ^z_z⁰ = 0.

Hence,

y⁰(x) = y(x−z)

x(z−y), z⁰(x) = z(y−x) x(z−y).

Since y⁰(x) < 0, the function y(x) is strictly decreasing. Since y(x₁) > x₁ (see sub-case x = x₁), there existsx₂ > x₁ such thaty(x₂) = x₂,y(x) > xforx₁ < x < x₂ andy(x) < x forx > x₂. Taking into account thaty≥x, it follows thatx₁ < x≤x₂. On the other hand, we see thatz⁰(x) >0forx₁ < x < x₂. Consequently, the functionz(x)is strictly increasing, and hencez(x) > z(x₁) = y(x₁) > y(x). Finally, we conclude thatx < y < z forx ∈ (x₁, x₂), andx=y < zforx=x₂.

B. Casep < 0. DenoteS = ^a+b+c₃ andR = ^ap+b₃^p+cp1_p

. Taking into account that x+y+z = 3S, x^p+y^p+z^p = 3R^p,

from0< x≤y≤zandx < zwe getx < S and3^1pR < x < R. Let h= (y+z)

y^p +z^p 2

⁻¹_p

−2.

By the AM-GM Inequality, we have h≥2√

yz 1

√yz −2 = 0,

with equality if and only ify=z. Writing nowhas a function ofx, h(x) = (3S−x)

3R^p −x^p 2

⁻¹_p

−2, from

h⁰(x) = 3R^p 2

3R^p −x^p 2

^−1−p_p "

S x

R x

−p

−1

#

>0 it follows that h(x) is strictly increasing. Since h(x) ≥ 0 andh

3^1pR

= −2, the equation h(x) = 0has a unique rootx₁ andx≥x₁ >3^1pR.

Sub-casex=x₁. Sincef(x) =f(x₁) = 0, andf = 0impliesy=z, we have0< x < y =z.

Sub-case x > x₁. We have h(x) > 0 and y < z. Consider now that y and z depend on x.

From x+y(x) +z(x) = 3S and x^p +y(x)^p +z(x)^p = 3R^p, we get 1 +y⁰ +z⁰ = 0 and x^p−1+y^p−1y⁰+z^p−1z⁰ = 0, and hence

y⁰(x) = x^p−1−z^p−1

z^p−1−y^p−1, z⁰(x) = x^p−1−y^p−1 y^p−1−z^p−1.

Since y⁰(x) > 0, the function y(x) is strictly decreasing. Since y(x₁) > x₁ (see sub-case x=x₁), there existsx₂ > x₁such thaty(x₂) =x₂,y(x)> xforx₁ < x < x₂, andy(x)< xfor x > x₂. The conditiony≥xyieldsx₁ < x≤x₂. We see now thatz⁰(x)>0forx₁ < x < x₂. Consequently, the functionz(x)is strictly increasing, and hencez(x)> z(x₁) =y(x₁)> y(x).

Finally, we havex < y < zforx∈(x₁, x₂)andx=y < zforx=x₂. C. Casep > 1. DenotingS = ^a+b+c₃ andR = ^ap+b₃^p+cp1_p

yields x+y+z = 3S, x^p+y^p+z^p = 3R^p.

By Jensen’s inequality applied to the convex functiong(u) = u^p, we have R > S, and hence x < S < R. Let

h= 2

y+z

y^p+z^p 2

¹_p

−1.

By Jensen’s Inequality, we geth≥0, with equality if only ify=z. From h(x) = 2

3S−x

3R^p−x^p 2

¹_p

−1 and

h⁰(x) = 3 (3S−x)²

3R^p−x^p 2

^1−p_p

(R^p−Sx^p−1)>0,

it follows that the functionh(x)is strictly increasing, andh(x)≥0impliesx≥x₁. In the case h(0) ≥0we havex₁ = 0, and in the caseh(0)<0we havex₁ >0andh(x₁) = 0.

Sub-casex=x₁. Ifh(0) ≥0, then0 =x₁ < y(x₁)≤z(x₁). Ifh(0) <0, thenh(x₁) = 0, and sinceh= 0impliesy=z, we have0< x₁ < y(x₁) =z(x₁).

Sub-casex > x1. Sinceh(x) is strictly increasing, for x > x1 we have h(x) > h(x1) ≥ 0, henceh(x)>0andy < z. Fromx+y(x) +z(x) = 3Sandx^p+y^p(x) +z^p(x) = 3R^p, we get

y⁰(x) = x^p−1−z^p−1

z^p−1−y^p−1, z⁰(x) = y^p−1−x^p−1 z^p−1−y^p−1.

Sincey⁰(x) < 0, the functiony(x) is strictly decreasing. Taking account of y(x₁) > x₁ (see sub-case x = x₁), there exists x₂ > x₁ such that y(x₂) = x₂, y(x) > x for x₁ < x < x₂, and y(x) < x for x > x₂. The condition y ≥ x implies x₁ < x ≤ x₂. We see now that z⁰(x) > 0 forx₁ < x < x₂. Consequently, the functionz(x)is strictly increasing, and hence z(x) > z(x₁) ≥ y(x₁) > y(x). Finally, we conclude that x < y < z for x ∈ (x₁, x₂), and

x=y < zforx=x₂.

Proof of Proposition 1.2. Consider the function

F(x) =f(x) +f(y(x)) +f(z(x))

defined onx∈ [x₁, x₂]. We claim thatF(x)is minimal forx=x₁ and is maximal forx =x₂. If this assertion is true, then by Lemma 1.1 it follows that:

(a) F(x) is minimal for 0 < x = y < z in the case p ≤ 0, or for either x = 0 or 0< x < y=z in the casep >1;

(b) F(x)is maximal for0< x=y < z.

In order to prove the claim, assume thatx∈(x₁, x₂). By Lemma 1.1, we have0< x < y <

z. From

x+y(x) +z(x) =a+b+c and x^p+y^p(x) +z^p(x) =a^p+b^p+c^p, we get

y⁰+z⁰ =−1, y^p−1y⁰+z^p−1z⁰ =−x^p−1, whence

y⁰ = x^p−1−z^p−1

z^p−1−y^p−1, z⁰ = x^p−1 −y^p−1 y^p−1−z^p−1. It is easy to check that this result is also valid forp= 0. We have

F⁰(x) =f⁰(x) +y⁰f⁰(y) +z⁰f⁰(z) and

F⁰(x)

(x^p−1−y^p−1)(x^p−1−z^p−1)

= g(x^p−1)

(x^p−1−y^p−1)(x^p−1−z^p−1) + g(y^p−1)

(y^p−1−z^p−1)(y^p−1−x^p−1) + g(z^p−1)

(z^p−1−x^p−1)(z^p−1−y^p−1). Sinceg is strictly convex, the right hand side is positive. On the other hand,

(x^p−1−y^p−1)(x^p−1−z^p−1)>0.

These results imply F⁰(x) > 0. Consequently, the function F(x) is strictly increasing for x∈ (x₁, x₂). Excepting the trivial case whenp >1,x₁ = 0andlim

u→0f(u) =−∞, the function

F(x)is continuous on[x₁, x₂], and hence is minimal only forx= x₁, and is maximal only for

x=x₂.

Proof of Theorem 1.3. We will consider two cases.

Casep∈(−∞,0]∪(1,∞). Excepting the trivial case whenp >1,x₁ = 0andlim

u→0f(u) =−∞, the functionFn(x1, x2, . . . , xn)attains its minimum and maximum values, and the conclusion follows from Proposition 1.2 above, via contradiction. For example, let us consider the casep≤ 0. In order to prove thatF_nis maximal for0< x₁ =x₂ =· · ·=x_n−1 ≤x_n, we assume, for the sake of contradiction, thatF_nattains its maximum at(b₁, b₂, . . . , b_n)withb₁ ≤ b₂ ≤ · · · ≤ b_n andb₁ < bn−1. Letx₁, xn−1, x_nbe positive numbers such thatx₁+xn−1+x_n=b₁+bn−1+b_n andx^p₁+x^p_n−1+x^p_n =b^p₁+b^p_n−1+b^p_n. According to Proposition 1.2, the expression

F₃(x₁, xn−1, x_n) = f(x₁) +f(xn−1) +f(x_n)

is maximal only for x₁ = x_n−1 < x_n, which contradicts the assumption that F_n attains its maximum at(b₁, b₂, . . . , b_n)withb₁ < b_n−1.

Casep ∈ (0,1). This case reduces to the casep > 1, replacing each of thea_i by a

1 p

i , each of thex_i byx

1 p

i , and thenpby ¹_p. Thus, we obtain the sufficient condition thath(x) =xf⁰ x^1−p1 to be strictly convex on(0,∞). We claim that this condition is equivalent to the condition that g(x) = f⁰

x^p−11

to be strictly convex on(0,∞). Actually, for our proof, it suffices to show that ifg(x)is strictly convex on (0,∞), thenh(x)is strictly convex on (0,∞). To show this, we see thatg _x¹

= ¹_xh(x). Sinceg(x)is strictly convex on(0,∞), by Jensen’s inequality we have

ug 1

x

+vg 1

y

>(u+v)g ^u

x +^v_y u+v

for anyx, y, u, v >0withx6=y. This inequality is equivalent to u

xh(x) + v

yh(y)>

u x+ v

y

h u+v

u x + ^v_y

! .

Substitutingu=txandv = (1−t)y, wheret∈(0,1), reduces the inequality to th(x) + (1−t)h(y)> h(tx+ (1−t)y),

which shows us thath(x)is strictly convex on(0,∞).

Proof of Corollary 1.8. We will apply Theorem 1.3 to the functionf(u) = plnu. We see that

u→0limf(u) =−∞forp >0, and f⁰(u) = p

u, g(x) =f⁰ x^p−11

=px^1−p1 , g⁰⁰(x) = p²

(1−p)²x^2p−11−p.

Sinceg⁰⁰(x) >0forx > 0, the functiong(x)is strictly convex on(0,∞), and the conclusion

follows by Theorem 1.3.

Proof of Corollary 1.9. We will apply Theorem 1.3 to the function f(u) =q(q−1)(q−p)u^q.

Forp > 0, it is easy to check that either f(u)is continuous at u = 0(in the case q > 0) or

u→0limf(u) =−∞(in the caseq <0). We have

f⁰(u) =q²(q−1)(q−p)u^q−1

and

g(x) = f⁰ x^p−11

=q²(q−1)(q−p)x^p−1q−1, g⁰⁰(x) = q²(q−1)²(q−p)²

(p−1)² x^2p−11−p.

Sinceg⁰⁰(x) >0forx > 0, the functiong(x)is strictly convex on(0,∞), and the conclusion

follows by Theorem 1.3.

3. APPLICATIONS

Proposition 3.1. Letx, y, zbe non-negative real numbers such thatx+y+z = 2. Ifr₀ ≤r≤3, wherer₀ = _{ln 3−ln 2}^{ln 2} ≈1.71, then

x^r(y+z) +y^r(z+x) +z^r(x+y)≤2.

Proof. Rewrite the inequality in the homogeneous form

x^r+1+y^r+1+z^r+1+ 2

x+y+z 2

r+1

≥(x+y+z)(x^r+y^r+z^r), and apply Corollary 1.9 (casep=randq=r+ 1):

If0≤x≤y≤zsuch that

x+y+z=constant and x^r+y^r+z^r=constant,

then the sumx^r+1+y^r+1+z^r+1is minimal when eitherx= 0or0< x≤y=z.

Casex= 0. The initial inequality becomes

yz(y^r−1+z^r−1)≤2,

wherey+z = 2. Since0< r−1≤2, by the Power Mean inequality we have y^r−1+z^r−1

2 ≤

y²+z² 2

^r−1₂ . Thus, it suffices to show that

yz

y²+z² 2

^r−1₂

≤1.

Taking account of

y²+z²

2 = 2(y²+z²)

(y+z)² ≥1 and r−1 2 ≤1, we have

1−yz

y²+z² 2

^r−1₂

≥1−yz

y² +z² 2

= (y+z)⁴

16 − yz(y²+z²) 2

= (y−z)⁴ 16 ≥0.

Case 0 < x ≤ y = z. In the homogeneous inequality we may leave aside the constraint x+y+z = 2, and considery =z = 1,0< x≤1. The inequality reduces to

1 + x

2 r+1

−x^r−x−1≥0.

Since 1 + ^x₂r+1

is increasing andx^ris decreasing in respect tor, it suffices to considerr=r0. Let

f(x) = 1 + x

2 r0+1

−x^r0 −x−1.

We have

f⁰(x) = r₀+ 1 2

1 + x

2 r0

−r₀x^r0−1−1, 1

r₀f⁰⁰(x) = r₀+ 1 4

1 + x

2 r0

− r₀−1 x^2−r0 . Sincef⁰⁰(x)is strictly increasing on(0,1],f⁰⁰(0₊) =−∞and

1

r₀f⁰⁰(1) = r₀+ 1 4

3 2

r0

−r0+ 1

= r₀+ 1

2 −r₀+ 1 = 3−r₀ 2 >0,

there existsx₁ ∈ (0,1)such that f⁰⁰(x₁) = 0, f⁰⁰(x) < 0 forx ∈ (0, x₁), and f⁰⁰(x) > 0for x ∈ (x₁,1]. Therefore, the function f⁰(x) is strictly decreasing for x ∈ [0, x₁], and strictly increasing forx∈[x₁,1]. Since

f⁰(0) = r₀−1

2 >0 and f⁰(1) = r₀+ 1 2

3 2

r0

−2

= 0,

there exists x2 ∈ (0, x1)such that f⁰(x2) = 0, f⁰(x) > 0for x ∈ [0, x2), and f⁰(x) < 0for x∈(x₂,1). Thus, the functionf(x)is strictly increasing forx∈[0, x₂], and strictly decreasing forx∈[x₂,1]. Sincef(0) =f(1) = 0, it follows thatf(x)≥0for0< x≤1, establishing the desired result.

Forx ≤ y≤ z, equality occurs whenx = 0andy = z = 1. Moreover, forr =r₀, equality

holds again whenx=y=z = 1.

Proposition 3.2 ([12]). Letx, y, z be non-negative real numbers such thatxy+yz +zx = 3.

If1< r≤2, then

x^r(y+z) +y^r(z+x) +z^r(x+y)≥6.

Proof. Rewrite the inequality in the homogeneous form

x^r(y+z) +y^r(z+x) +z^r(x+y)≥6

xy+yz+zx 3

^r+1₂ .

For convenience, we may leave aside the constraintxy+yz+zx= 3. Using now the constraint x+y+z = 1, the inequality becomes

x^r(1−x) +y^r(1−y) +z^r(1−z)≥6

1−x² −y²−z² 6

^r+1₂ .

To prove it, we will apply Corollary 1.5 to the functionf(u) =−u^r(1−u)for0≤u≤1. We havef⁰(u) = −ru^r−1+ (r+ 1)u^r and

g(x) =f⁰(x) =−rx^r−1+ (r+ 1)x^r, g⁰⁰(x) =r(r−1)x^r−3[(r+ 1)x+ 2−r].

Since g⁰⁰(x) > 0 for x > 0, g(x) is strictly convex on [0,∞). According to Corollary 1.5, if 0 ≤ x ≤ y ≤ z such that x+y +z = 1 and x² +y² +z² = constant, then the sum f(x) +f(y) +f(z)is maximal for0≤x=y≤z.

Thus, we have only to prove the original inequality in the casex = y ≤ z. This means, to prove that0< x≤1≤yandx²+ 2xz = 3implies

x^r(x+z) +xz^r ≥3.

Letf(x) =x^r(x+z) +xz^r−3,withz = ^3−x_2x².

Differentiating the equationx²+ 2xz = 3yieldsz⁰ = ^−(x+z)_x . Then, f⁰(x) = (r+ 1)x^r+rx^r−1z+z^r+ (x^r+rxz^r−1)z⁰

= (x^r−1−z^r−1)[rx+ (r−1)z]≤0.

The functionf(x)is strictly decreasing on[0,1], and hencef(x) ≥ f(1) = 0for0 < x ≤ 1.

Equality occurs if and only ifx=y=z = 1.

Proposition 3.3 ([5]). Ifx₁, x₂, . . . , x_nare positive real numbers such that x₁+x₂+· · ·+x_n = 1

x₁ + 1

x₂ +· · ·+ 1 x_n, then

1

1 + (n−1)x₁ + 1

1 + (n−1)x₂ +· · ·+ 1

1 + (n−1)x_n ≥1.

Proof. We have to consider two cases.

Casen = 2. The inequality is verified as equality.

Casen≥3. Assume that0< x₁ ≤x₂ ≤ · · · ≤x_n, and then apply Corollary 1.6 to the function f(u) = _1+(n−1)u¹ foru >0. We havef⁰(u) = _[1+(n−1)u]⁻⁽ⁿ⁻¹⁾ 2 and

g(x) = f⁰ 1

√x

= −(n−1)x (√

x+n−1)², g⁰⁰(x) = 3(n−1)²

2√ x(√

x+n−1)⁴.

Sinceg⁰⁰(x) > 0, g(x)is strictly convex on(0,∞). According to Corollary 1.6, if 0 < x1 ≤ x₂ ≤ · · · ≤x_nsuch that

x1+x2 +· · ·+xn=constant and 1

x₁ + 1

x₂ +· · ·+ 1

x_n =constant,

then the sumf(x₁) +f(x₂) +· · ·+f(x_n)is minimal when0< x₁ ≤x₂ =x₃ =· · ·=x_n. Thus, we have to prove the inequality

1

1 + (n−1)x + n−1

1 + (n−1)y ≥1, under the constraints0< x≤1≤yand

x+ (n−1)y= 1

x +n−1 y . The last constraint is equivalent to

(n−1)(y−1) = y(1−x²) x(1 +y).

Since

1

1 + (n−1)x + n−1

1 + (n−1)y −1

= 1

1 + (n−1)x− 1

n + n−1

1 + (n−1)y − n−1 n

= (n−1)(1−x)

n[1 + (n−1)x] −(n−1)²(y−1) n[1 + (n−1)y]

= (n−1)(1−x)

n[1 + (n−1)x] − (n−1)y(1−x²) nx(1 +y)[1 + (n−1)y], we must show that

x(1 +y)[1 + (n−1)y]≥y(1 +x)[1 + (n−1)x], which reduces to

(y−x)[(n−1)xy−1]≥0.

Sincey−x≥0, we have still to prove that

(n−1)xy≥1.

Indeed, fromx+ (n−1)y= _x¹ + ⁿ⁻¹_y we getxy= ^y+(n−1)x_x+(n−1)y, and hence (n−1)xy−1 = n(n−2)x

x+ (n−1)y >0.

Forn ≥3, one has equality if and only ifx1 =x2 =· · ·=xn= 1.

Proposition 3.4 ([10]). Leta1, a2, . . . , anbe positive real numbers such thata1a2· · ·an= 1. If mis a positive integer satisfyingm≥n−1, then

a^m₁ +a^m₂ +· · ·+a^m_n + (m−1)n≥m 1

a₁ + 1

a₂ +· · ·+ 1 a_n

. Proof. Forn= 2(hencem ≥1), the inequality reduces to

a^m₁ +a^m₂ + 2m−2≥m(a₁+a₂).

We can prove it by summing the inequalitiesa^m₁ ≥1+m(a₁−1)anda^m₂ ≥1+m(a₂−1), which are straightforward consequences of Bernoulli’s inequality. Forn≥ 3, replacinga₁, a₂, . . . , a_n by _x¹

1,_x¹

2, . . . ,_x¹

n, respectively, we have to show that 1

x^m₁ + 1

x^m₂ +· · ·+ 1

x^m_n + (m−1)n ≥m(x1+x2+· · ·+xn)

forx₁x₂· · ·x_n = 1. Assume0< x₁ ≤x₂ ≤ · · · ≤x_nand apply Corollary 1.9 (casep= 0and q=−m):

If0< x₁ ≤x₂ ≤ · · · ≤x_nsuch that

x1+x2+· · ·+xn =constant and x₁x₂· · ·x_n = 1,

then the sum _x¹m 1

+ _x¹m 2

+· · ·+_x¹m

n is minimal when0< x₁ =x₂ =· · ·=x_n−1 ≤x_n.

Thus, it suffices to prove the inequality forx₁ = x₂ = · · · = xn−1 = x ≤ 1, x_n = yand xⁿ⁻¹y= 1, when it reduces to:

n−1 x^m + 1

y^m + (m−1)n≥m(n−1)x+my.

By the AM-GM inequality, we have n−1

x^m + (m−n+ 1) ≥ m

xⁿ⁻¹ =my.

Then, we have still to show that 1

y^m −1≥m(n−1)(x−1).

This inequality is equivalent to

x^mn−m−1−m(n−1)(x−1)≥0 and

(x−1)[(x^mn−m−1−1) + (x^mn−m−2−1) +· · ·+ (x−1)] ≥0.

The last inequality is clearly true. For n = 2 and m = 1, the inequality becomes equality.

Otherwise, equality occurs if and only ifa₁ =a₂ =· · ·=a_n = 1.

Proposition 3.5 ([6]). Letx₁, x₂, . . . , x_nbe non-negative real numbers such thatx₁+x₂+· · ·+ x_n=n. Ifkis a positive integer satisfying2≤k≤n+ 2, andr= _n−1^{n k−1}

−1, then x^k₁ +x^k₂+· · ·+x^k_n−n≥nr(1−x1x2· · ·xn).

Proof. Ifn = 2, then the inequality reduces tox^k₁ +x^k₂ −2 ≥ (2^k−2)x₁x₂. Fork = 2and k = 3, this inequality becomes equality, while fork = 4 it reduces to6x₁x₂(1−x₁x₂) ≥ 0, which is clearly true.

Consider now n ≥ 3 and 0 ≤ x₁ ≤ x₂ ≤ · · · ≤ x_n. Towards proving the inequality, we will apply Corollary 1.8 (case p = k > 0): If 0 ≤ x₁ ≤ x₂ ≤ · · · ≤ x_n such that x₁ +x₂ +· · ·+x_n = n andx^k₁ +x^k₂ +· · ·+x^k_n =constant, then the productx₁x₂· · ·x_n is minimal when eitherx₁ = 0or0< x₁ ≤x₂ =x₃ =· · ·=x_n.

Casex1 = 0. The inequality reduces to

x^k₂ +· · ·+x^k_n ≥ n^k (n−1)^k−1,

withx₂+· · ·+x_n =n, This inequality follows by applying Jensen’s inequality to the convex functionf(u) =u^k:

x^k₂+· · ·+x^k_n ≥(n−1)

x₂+· · ·+x_n n−1

k

.

Case0< x₁ ≤ x₂ =x₃ =· · ·= x_n. Denotingx₁ =xandx₂ =x₃ = · · ·=x_n =y, we have to prove that for0< x≤1≤yandx+ (n−1)y =n, the inequality holds:

x^k+ (n−1)y^k+nrxyⁿ⁻¹−n(r+ 1)≥0.

Write the inequality asf(x)≥0, where

f(x) =x^k+ (n−1)y^k+nrxyⁿ⁻¹−n(r+ 1), with y= n−x n−1. We see thatf(0) =f(1) = 0. Sincey⁰ = _n−1⁻¹ , we have

f⁰(x) = k(x^k−1−y^k−1) +nryⁿ⁻²(y−x)

= (y−x)[nryⁿ⁻²−k(y^k−2+y^k−3x+· · ·+x^k−2)]

= (y−x)yⁿ⁻²[nr−kg(x)],

where

g(x) = 1

y^n−k + x

y^n−k+1 +· · ·+x^k−2 yⁿ⁻².

Since the functiony(x) = ^n−x_n−1 is strictly decreasing, the functiong(x)is strictly increasing for 2≤k≤n. Fork =n+ 1, we have

g(x) =y+x+x²

y +· · ·+xⁿ⁻¹ yⁿ⁻²

= (n−2)x+n n−1 +x²

y +· · ·+xⁿ⁻¹ yⁿ⁻², and fork=n+ 2, we have

g(x) =y²+yx+x²+x³

y +· · ·+ xⁿ yⁿ⁻²

= (n²−3n+ 3)x²+n(n−3)x+n² (n−1)² +x³

y +· · ·+ xⁿ yⁿ⁻². Therefore, the functiong(x)is strictly increasing for2≤k ≤n+ 2, and the function

h(x) =nr−kg(x) is strictly decreasing. Note that

f⁰(x) = (y−x)yⁿ⁻²h(x).

We assert thath(0)>0andh(1)<0. If our claim is true, then there existsx1 ∈(0,1)such that h(x1) = 0,h(x)>0forx∈[0, x1), andh(x)<0forx∈(x1,1]. Consequently,f(x)is strictly increasing for x ∈ [0, x₁], and strictly decreasing for x ∈ [x₁,1]. Since f(0) = f(1) = 0, it follows thatf(x)≥0for0< x≤1, and the proof is completed.

In order to prove thath(0) > 0, we assume thath(0) ≤ 0. Then, h(x) < 0 forx ∈ (0,1), f⁰(x) < 0 for x ∈ (0,1), and f(x) is strictly decreasing for x ∈ [0,1], which contradicts f(0) = f(1). Also, ifh(1) ≥ 0, thenh(x) > 0 forx ∈ (0,1), f⁰(x) > 0forx ∈ (0,1), and f(x)is strictly increasing forx∈[0,1], which also contradictsf(0) = f(1).

Forn≥3andx₁ ≤x₂ ≤ · · · ≤x_n, equality occurs whenx₁ =x₂ =· · ·=x_n = 1, and also

whenx₁ = 0andx₂ =· · ·=x_n= _n−1ⁿ .

Remark 3.6. Fork= 2,k = 3andk = 4, we get the following nice inequalities:

(n−1)(x²₁+x²₂+· · ·+x²_n) +nx₁x₂· · ·x_n≥n², (n−1)²(x³₁+x³₂+· · ·+x³_n) +n(2n−1)x1x2· · ·xn ≥n³, (n−1)³(x⁴₁+x⁴₂+· · ·+x⁴_n) +n(3n²−3n+ 1)x₁x₂· · ·x_n ≥n⁴.

Remark 3.7. The inequality fork =nwas posted in 2004 on the Mathlinks Site - Inequalities Forum by Gabriel Dospinescu and C˘alin Popa.

Proposition 3.8 ([11]). Letx₁, x₂, . . . , x_nbe positive real numbers such that_x¹

1+_x¹

2+· · ·+_x¹

n = n. Then

x₁+x₂+· · ·+x_n−n ≤e_n−1(x₁x₂· · ·x_n−1), wheree_n−1 = 1 + _n−1^{1 n−1}

< e.

Proof. Replacing each of thex_i by _a¹

i, the statement becomes as follows:

Ifa₁, a₂, . . . , a_nare positive numbers such thata₁ +a₂ +· · ·+a_n=n, then a₁a₂· · ·a_n

1 a₁ + 1

a₂ +· · ·+ 1

a_n −n+en−1

≤en−1.

It is easy to check that the inequality holds for n = 2. Consider now n ≥ 3, assume that 0 < a₁ ≤ a₂ ≤ · · · ≤ a_n and apply Corollary 1.8 (casep = −1): If 0< a₁ ≤ a₂ ≤ · · · ≤ a_n such thata1+a2+· · ·+an =nand _a¹

1 +_a¹

2+· · ·+_a¹

n =constant, then the producta1a2· · ·an

is maximal when0< a₁ ≤a₂ =a₃ =· · ·=a_n.

Denotinga₁ = xanda₂ =a₃ =· · · =a_n =y, we have to prove that for 0< x ≤1≤ y <

n

n−1 andx+ (n−1)y=n, the inequality holds:

yⁿ⁻¹+ (n−1)xyⁿ⁻²−(n−e_n−1)xyⁿ⁻¹ ≤e_n−1. Letting

f(x) =yⁿ⁻¹+ (n−1)xyⁿ⁻²−(n−en−1)xyⁿ⁻¹−en−1, with y= n−x

n−1,

we must show thatf(x)≤0for0< x≤1. We see thatf(0) =f(1) = 0. Sincey⁰ = _n−1⁻¹ , we

have f⁰(x)

yⁿ⁻³ = (y−x)[n−2−(n−en−1)y] = (y−x)h(x), where

h(x) = n−2−(n−en−1)n−x n−1 is a linear increasing function.

Let us show that h(0) < 0 and h(1) > 0. If h(0) ≥ 0, then h(x) > 0 for x ∈ (0,1), hencef⁰(x) >0forx∈ (0,1), andf(x)is strictly increasing forx∈ [0,1], which contradicts f(0) =f(1). Also,h(1) =en−1−2>0.

Fromh(0) < 0and h(1) > 0, it follows that there existsx₁ ∈ (0,1)such that h(x₁) = 0, h(x)<0forx∈[0, x₁), andh(x)>0forx∈(x₁,1]. Consequently,f(x)is strictly decreasing forx ∈ [0, x₁], and strictly increasing for x ∈ [x₁,1]. Since f(0) = f(1) = 0, it follows that f(x)≤0for0≤x≤1.

Forn ≥3, equality occurs whenx₁ =x₂ =· · ·=x_n= 1.

Proposition 3.9 ([9]). Ifx1, x2, . . . , xnare positive real numbers, then xⁿ₁ +xⁿ₂ +· · ·+xⁿ_n+n(n−1)x₁x₂· · ·x_n

≥x₁x₂· · ·x_n(x₁+x₂ +· · ·+x_n) 1

x₁ + 1

x₂ +· · ·+ 1 x_n

. Proof. For n = 2, one has equality. Assume now that n ≥ 3, 0 < x₁ ≤ x₂ ≤ · · · ≤ x_nand apply Corollary 1.9 (casep= 0): If0< x₁ ≤x₂ ≤ · · · ≤x_nsuch that

x₁+x₂+· · ·+x_n =constant and x₁x₂· · ·x_n =constant, then the sumxⁿ₁ +xⁿ₂ +· · ·+xⁿ_nis minimal and the sum _x¹

1 +_x¹

2 +· · ·+_x¹

n is maximal when 0< x₁ ≤x₂ =x₃ =· · ·=x_n.

Thus, it suffices to prove the inequality for0 < x₁ ≤ 1andx₂ = x₃ = · · · = x_n = 1. The inequality becomes

xⁿ₁ + (n−2)x₁ ≥(n−1)x²₁,

and is equivalent to

x₁(x₁ −1)[(xⁿ⁻²₁ −1) + (xⁿ⁻³₁ −1) +· · ·+ (x₁−1)]≥0,

which is clearly true. Forn ≥3, equality occurs if and only ifx₁ =x₂ =· · ·=x_n. Proposition 3.10 ([14]). Ifx1, x2, . . . , xnare non-negative real numbers, then

(n−1)(xⁿ₁ +xⁿ₂ +· · ·+xⁿ_n) +nx₁x₂· · ·x_n

≥(x1+x2+· · ·+xn)(xⁿ⁻¹₁ +xⁿ⁻¹₂ +· · ·+xⁿ⁻¹_n ).

Proof. For n = 2, one has equality. For n ≥ 3, assume that 0 ≤ x₁ ≤ x₂ ≤ · · · ≤ x_n and apply Corollary 1.9 (case p = n and q = n − 1) and Corollary 1.8 (case p = n): If 0≤x₁ ≤x₂ ≤ · · · ≤x_nsuch that

x₁+x₂+· · ·+x_n=constant and xⁿ₁ +xⁿ₂ +· · ·+xⁿ_n=constant,

then the sumxⁿ⁻¹₁ +xⁿ⁻¹₂ +· · ·+xⁿ⁻¹_n is maximal and the productx₁x₂· · ·x_nis minimal when eitherx₁ = 0or0< x₁ ≤x₂ =x₃ =· · ·=x_n.

So, it suffices to consider the casesx1 = 0and0< x1 ≤x2 =x3 =· · ·=xn. Casex₁ = 0. The inequality reduces to

(n−1)(xⁿ₂ +· · ·+xⁿ_n)≥(x₂+· · ·+x_n)(xⁿ⁻¹₂ +· · ·+xⁿ⁻¹_n ), which immediately follows by Chebyshev’s inequality.

Case0< x₁ ≤ x₂ = x₃ = · · ·= x_n. Settingx₂ =x₃ =· · ·= x_n = 1, the inequality reduces to:

(n−2)xⁿ₁ +x₁ ≥(n−1)xⁿ⁻¹₁ . Rewriting this inequality as

x₁(x₁−1)[xⁿ⁻³₁ (x₁−1) +xⁿ⁻⁴₁ (x²₁−1) +· · ·+ (xⁿ⁻²₁ −1)]≥0,

we see that it is clearly true. For n ≥ 3 and x₁ ≤ x₂ ≤ · · · ≤ x_n equality occurs when x₁ =x₂ =· · ·=x_n, and forx₁ = 0andx₂ =· · ·=x_n. Proposition 3.11 ([8]). Ifx₁, x₂, . . . , x_nare positive real numbers, then

(x₁+x₂+· · ·+x_n−n) 1

x₁ + 1

x₂ +· · ·+ 1 x_n −n

+x₁x₂· · ·x_n+ 1

x₁x₂· · ·x_n ≥2.

Proof. Forn= 2, the inequality reduces to

(1−x₁)²(1−x₂)² x₁x₂ ≥0.

For n ≥ 3, assume that0 < x1 ≤ x2 ≤ · · · ≤ xn. Since the inequality preserves its form by replacing each numberxi with _x¹

i, we may considerx1x2· · ·xn ≥ 1. So, by the AM-GM inequality we get

x₁+x₂+· · ·+x_n−n≥n√ⁿ

x₁x₂· · ·x_n−n≥0,

and we may apply Corollary 1.9 (casep = 0andq = −1): If0 ≤ x₁ ≤ x₂ ≤ · · · ≤ x_nsuch that

x1+x2+· · ·+xn =constant and x₁x₂· · ·x_n =constant, then the sum _x¹

1 + _x¹

2 +· · ·+ _x¹

n is minimal when0< x₁ =x₂ =· · ·=xn−1 ≤x_n.