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let $m,n$ be integers, show that if $ n>m\geq 0 $ :

$$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3} {2}\left(\frac{1}{\sqrt{3}}\right)^{n-m}$$

where real $x,y,z > 0 $ and $xy + yz + zx = 1$

Thank you for your help .

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  • $\begingroup$ according to the RHS it's seems as a power of geometric sequence , but i can't up to it $\endgroup$ May 31, 2015 at 15:23
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    $\begingroup$ If I am not wrong you have to edit your question,Zeraoulia. When x = y = z one has x $\geq$ $\frac{1}{\sqrt3}$ (so not for all real) $\endgroup$
    – Piquito
    May 31, 2015 at 15:39
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    $\begingroup$ "Tim Carson: I didn't see at all that condition. Sorry (now I simply accept the statement without verification) $\endgroup$
    – Piquito
    Jun 1, 2015 at 13:47
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    $\begingroup$ Problem 111 from mathproblems-ks.com/?wpfb_dl=59 ... posted here before the July 15 deadline for submitting answers to the journal. $\endgroup$
    – GEdgar
    Sep 16, 2015 at 15:37
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    $\begingroup$ Please see correction of this problem in the last issue of Mathproblems journal. The condition on n and m should be changed. $\endgroup$
    – MotiLevy
    Oct 17, 2016 at 18:39

3 Answers 3

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Despite the attempts in the other answers and comments, and the two ``proofs'' given there, the inequality does not hold in general. We give two proofs for its failure:

First Proof (involving huge numbers): Set $x=3/7$, $y=4/7$, $z=37/49$. Then $xy+yz+zx=1$. For $m=7$, $n=8$ set $A=\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}$ and $B=\frac{3}{2}(\frac{1}{\sqrt{3}})^{n-m}$. Then \begin{equation} A^2-B^2=-\frac{5464419604082977128654242570410694589510448147711713}{929486260504473222256638487813651283882185173994428050}<0. \end{equation}

Second proof (more conceptual): Suppose that $0<x<y<z$. Then, with $n=m+1$, \begin{equation} \lim_{m\to\infty} (\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}) = \lim_{m\to\infty} (\frac{x}{1+(y/x)^m}+\frac{y}{1+(z/y)^m}+\frac{z}{1+(x/z)^m})=z \end{equation} Now set $x=\frac{1}{3}$, $y=\frac{2}{3}$, $z=\frac{7}{9}$. Then $xy+yz+zx=1$, but $z<\frac{3}{2}\frac{1}{\sqrt{3}}$. So the stated inequality does not hold for all sufficiently large $m$.

Remark: The same question was asked here at MathOverflow, and I gave a similar counterexample there.

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This is a sketch with some details I didn't want to fill in but they'll check out. Let $n>m\geq 0$ be fixed and $$f(x,y,z) = \frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}$$ we'll try to find the minimum under $xz+zy+yx=1$ using Lagrange multipliers. Let $$ g(x,y,z) = xz + zy+ yx$$ Following the idea of the multipliers, we try to solve $$ \nabla f = \lambda \nabla g$$ We see the first relationship as $$ \frac{(n-m) x^{n-1}x^m+ n y^m x^{n-1}}{(x^m + y^m)^2}-m\frac{ x^{m-1} z^n }{(z^m + x^m)^2} = \lambda(z+y) $$ It's not so easy to see, but since all 3 equations are the same having $x,y,z$ mixed up, it means we must have $x=y=z= \frac{1}{\sqrt{3}}$ for the three equations to hold with $$ \lambda = \frac{ n-m}{4} \left ( \frac{1}{\sqrt{3}} \right) ^{n-m-2} $$ Also, you'll see that this value is a minimum by checking the Hessian(or by a clever intuition), use $g(x,y,z)=1$ ,$x,y,z,>0$ and $n >m$. Thus $$f(x,y,z) \Big | _g \geq \min_{x,y,z} f(x,y,z) \Big | _g = f \left(\frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}} \right)= \frac{3}{2} \left ( \frac{1}{\sqrt{3}} \right)^{n-m} $$ Edit: It occurred to me I should say something about the max of $f(x,y,z) \big| _g$, you'll see it occurs at infinity since we may take $ y = \epsilon \approx 0$ then $$xz \approx 1 \implies x \approx \frac{1}{z}$$ So as $x \to \infty$ we see that $$f(x,y,z) \Big |_g = \mathcal{O} (x^{n-m}) \to \infty $$

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By the conditions, we have: $$y =\frac{1-xz}{x+z}\ge0\quad\longrightarrow \quad xz\le1, z \le 1/x$$ Now, without loss of generality, assume $z\ge y\ge x$. We therefore have that the expression $f(x,y,z)$ in question satisfies: $$f(x,y,z) \ge \frac{3}{2} \frac{x^n}{z^m}\ge \frac{3}{2}x^{n-m}$$

Now it remains to be shown that $x\ge 1/\sqrt{3}$. This is true since the minimal distance from the origin to the hyperboloid is exactly $1$, so that: $$x^2+y^2+z^2 \ge 1 \quad\longrightarrow \quad 3x^2 \ge 1$$ Thus completing the proof.

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  • $\begingroup$ @Macavity - if $z\ge x$, then all numerators are larger than $x^n$, and all denominators are smaller than $2z^m$. $\endgroup$ Jun 1, 2015 at 8:58
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    $\begingroup$ Now in the edited version, how can you say $\dfrac32 \dfrac{x^n}{z^m} \ge \dfrac32 x^{n-m}$? That can hold only with your earlier assumption, not with this one. $$ $$Similarly, it is no longer true that $x \ge \frac1{\sqrt3}$, in fact $x \le \frac1{\sqrt3}$ now, as $1 = xy+yz+zx \ge 3x^2$. $\endgroup$
    – Macavity
    Jun 1, 2015 at 13:44
  • $\begingroup$ @Macavity is right. With the above reasoning you can only get: $\frac{x^n}{z^m} = x^n\left(\frac{1}{z}\right)^m \ge x^nx^m = x^{n+m}$. Plus, not minus. $\endgroup$ Sep 14, 2015 at 17:12

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