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

$\displaystyle \frac{x^{y+z}}{(y+z)^2}+\frac{y^{z+x}}{(z+x)^2}+\frac{z^{x+y}}{(x+y)^2}\geq \frac{3}{4}$

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

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## Comments

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

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

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

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

$\geq 3\frac{S}{3}$

$\geq 3\frac{{(\frac{S}{3})}^{\frac{2S}{3}}}{{\frac{2S}{3}}^{2}}$

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

Now, we need to find the minimum value of ${(\frac{S}{3})}^{\frac{2S-6}{3}}$. Taking derivative of it w.r.t. S and equating to 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 ${(\frac{S}{3})}^{\frac{2S-6}{3}} = 0$, which is not possible as $S > 0$

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

After some bashing, this can be written as,

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

Hence, $S = 3$.

Substituting in 1,

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

which is just

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

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

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

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nice solution @Kartik Sharma

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

$\dfrac {\dfrac {x^{y+z}}{(y+z)^2} + \dfrac {y^{z+x}}{(z+x)^2} + \dfrac {z^{x+y}}{(x+y)^2} } {3} \ge \sqrt [3] {\left( \dfrac {x^{y+z}} {(y+z)^2} \right) \left( \dfrac {y^{z+x}}{(z+x)^2} \right) \left( \dfrac {z^{x+y}}{(x+y)^2} \right) }$

$\dfrac {x^{y+z}}{(y+z)^2} + \dfrac {y^{z+x}}{(z+x)^2} + \dfrac {z^{x+y}}{(x+y)^2} \ge 3 \sqrt [3] {\dfrac {x^{y+z}y^{z+x} z^{x+y}} {(x+y)^2(y+z)^2(z+x)^2} }$

$\dfrac {x^{y+z}}{(y+z)^2} + \dfrac {y^{z+x}}{(z+x)^2} + \dfrac {z^{x+y}}{(x+y)^2} \ge \dfrac {3x^{\frac {1}{3} (y+z)}y^{\frac {1}{3} (z+x)} z^{\frac {1}{3} (x+y)}} {[(x+y)(y+z)(z+x)]^\frac{2}{3}}$

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

$\Rightarrow \dfrac {x^{y+z}}{(y+z)^2} + \dfrac {y^{z+x}}{(z+x)^2} + \dfrac {z^{x+y}}{(x+y)^2} \ge \dfrac {3\dot{} 1\dot{}1\dot{}1} {[2\dot{}2\dot{}2]^\frac{2}{3}} = \dfrac {3}{4}$

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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 $x=y=z=1$.

@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 $x=y=z$, it's pretty non-trivial to prove that the minimum is when $x=y=z=1$. I'm guessing that this inequality cannot be proved using elementary techniques.

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

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

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Let $s=x+y+z$ and $f(x)=\frac{x^{s-x}}{(s-x)^2}$. Confirm that $f$ is convex on $0<x<s$. 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

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@Deepanshu Gupta @Sandeep Bhardwaj @Sanjeet Raria @Pranjal Jain @abdulrahman khaled

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@Daniel Liu Can you help us with this problem?

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

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

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Check now

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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..

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