Solving Inequality Problem

Let x,y,zx, y, z be positive real numbers . Prove that:

xy+z(y+z)2+yz+x(z+x)2+zx+y(x+y)234 \displaystyle \frac{x^{y+z}}{(y+z)^2}+\frac{y^{z+x}}{(z+x)^2}+\frac{z^{x+y}}{(x+y)^2}\geq \frac{3}{4}

I always confused how can I solve inequality with variable-power on it. Appreciate your help to solve this problem for sure :)

Note by Natasha Astriani
5 years, 2 months ago

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I thought of this on my own but I saw that Cody Johnson has already shared it.

So, showing that ff is convex for the interval [0,S] ( Take second-order derivative or the 'definition' method)

Applying Jensen's inequality for f(a)=aSaSa2f(a) = \frac{{a}^{S-a}}{{S-a}^{2}},

f(x)+f(y)+f(z)3x+y+z3f(x) + f(y) + f(z) \geq 3\frac{x+y+z}{3}

3S3\geq 3\frac{S}{3}

3(S3)2S32S32\geq 3\frac{{(\frac{S}{3})}^{\frac{2S}{3}}}{{\frac{2S}{3}}^{2}}

3(S3)2S634\geq 3\frac{{(\frac{S}{3})}^{\frac{2S-6}{3}}}{4} --------- 1

Now, we need to find the minimum value of (S3)2S63{(\frac{S}{3})}^{\frac{2S-6}{3}}. Taking derivative of it w.r.t. S and equating to 0,

(S3)2S63(2S63S+23ln(S3))=0{(\frac{S}{3})}^{\frac{2S-6}{3}}(\frac{2S-6}{3S} + \frac{2}{3}ln(\frac{S}{3})) = 0 [ln(b) is the logarithm base e of b]

So, either (S3)2S63=0{(\frac{S}{3})}^{\frac{2S-6}{3}} = 0, which is not possible as S>0S > 0

or 2S63S+23ln(S3)=0\frac{2S-6}{3S} + \frac{2}{3}ln(\frac{S}{3}) = 0

After some bashing, this can be written as,

S(ln(S3))=3Sec=ecS(ln(\frac{S}{3})) = 3 - S \Rightarrow {e}^{c} = \frac{e}{c} [for c = 3S\frac{3}{S}], which is only possible for c = 1.

Hence, S=3S = 3.

Substituting in 1,

f(x)+f(y)+f(z)3(33)23634f(x) + f(y) + f(z) \geq 3\frac{{(\frac{3}{3})}^{\frac{2*3-6}{3}}}{4}

which is just

f(x)+f(y)+f(z)34f(x) + f(y) + f(z) \geq \frac{3}{4}

Kartik Sharma - 4 years, 9 months ago

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@Cody Johnson I hope I didn't copy and if I did, then I really didn't mean it!

Kartik Sharma - 4 years, 9 months ago

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Wow, this is so obvious now that I read your solution... :(

Daniel Liu - 4 years, 9 months ago

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nice solution @Kartik Sharma

Mardokay Mosazghi - 4 years, 9 months ago

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We know that AMGMAM \ge GM and they are equal when x=y=z=1x=y=z=1. Therefore,

xy+z(y+z)2+yz+x(z+x)2+zx+y(x+y)23(xy+z(y+z)2)(yz+x(z+x)2)(zx+y(x+y)2)3\dfrac {\dfrac {x^{y+z}}{(y+z)^2} + \dfrac {y^{z+x}}{(z+x)^2} + \dfrac {z^{x+y}}{(x+y)^2} } {3} \ge \sqrt [3] {\left( \dfrac {x^{y+z}} {(y+z)^2} \right) \left( \dfrac {y^{z+x}}{(z+x)^2} \right) \left( \dfrac {z^{x+y}}{(x+y)^2} \right) }

xy+z(y+z)2+yz+x(z+x)2+zx+y(x+y)23xy+zyz+xzx+y(x+y)2(y+z)2(z+x)23\dfrac {x^{y+z}}{(y+z)^2} + \dfrac {y^{z+x}}{(z+x)^2} + \dfrac {z^{x+y}}{(x+y)^2} \ge 3 \sqrt [3] {\dfrac {x^{y+z}y^{z+x} z^{x+y}} {(x+y)^2(y+z)^2(z+x)^2} }

xy+z(y+z)2+yz+x(z+x)2+zx+y(x+y)23x13(y+z)y13(z+x)z13(x+y)[(x+y)(y+z)(z+x)]23\dfrac {x^{y+z}}{(y+z)^2} + \dfrac {y^{z+x}}{(z+x)^2} + \dfrac {z^{x+y}}{(x+y)^2} \ge \dfrac {3x^{\frac {1}{3} (y+z)}y^{\frac {1}{3} (z+x)} z^{\frac {1}{3} (x+y)}} {[(x+y)(y+z)(z+x)]^\frac{2}{3}}

Equality occurs when x=y=z=1x=y=z=1:

xy+z(y+z)2+yz+x(z+x)2+zx+y(x+y)23˙1˙1˙1[2˙2˙2]23=34\Rightarrow \dfrac {x^{y+z}}{(y+z)^2} + \dfrac {y^{z+x}}{(z+x)^2} + \dfrac {z^{x+y}}{(x+y)^2} \ge \dfrac {3\dot{} 1\dot{}1\dot{}1} {[2\dot{}2\dot{}2]^\frac{2}{3}} = \dfrac {3}{4}

Chew-Seong Cheong - 4 years, 9 months ago

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I don't see how you proved the inequality... You just manipulated it a bit and then claimed that equality happens at x=y=z=1x=y=z=1.

@Mardokay Mosazghi I'll try to solve it. I don't usually tackle problems with variable exponents like this one though.

EDIT: I honestly have no idea how to prove this. Even when assuming x=y=zx=y=z, it's pretty non-trivial to prove that the minimum is when x=y=z=1x=y=z=1. I'm guessing that this inequality cannot be proved using elementary techniques.

Daniel Liu - 4 years, 9 months ago

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Uhm Let a = x + z, b = y + z, and c = x + y?

Rindell Mabunga - 4 years, 9 months ago

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@Rindell Mabunga That's a good idea, to substitute something in. However your substitution (called Ravi's substitution) is only applicable if x, y, and z are the sides of a triangle.

Daniel Liu - 4 years, 9 months ago

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Daniel, you are right. I thought I was using the theorem. Can anyone help?

Chew-Seong Cheong - 4 years, 9 months ago

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Let s=x+y+zs=x+y+z and f(x)=xsx(sx)2f(x)=\frac{x^{s-x}}{(s-x)^2}. Confirm that ff is convex on 0<x<s0<x<s. Apply Jensen's inequality.

I'm not sure if this is correct, but I'm pretty darn sure… can't confirm because I'm on my phone

Cody Johnson - 4 years, 9 months ago

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@Daniel Liu Can you help us with this problem?

Mardokay Mosazghi - 4 years, 9 months ago

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Put x =1 and then 0 it means limit are satisfying. Rest I don't know

Rohit Singh - 4 years, 9 months ago

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why do you do this for x only, please elaborate.

Mardokay Mosazghi - 4 years, 9 months ago

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Check now

Rohit Singh - 4 years, 9 months ago

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Yes its easy u put all x.y.z 1 to prove equality and gv counter example by taking all zero which is not a poitive integer. Hence inequality is not proved on other then positive.. And as it is a polynomial it will form a ring hence close under both operation so inequality is proved..

Rohit Singh - 4 years, 9 months ago

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