Given that $xyz=1$, prove that $$\frac{x}{1+x^4}+\frac{y}{1+y^4}+\frac{z}{1+z^4}\le\frac{3}{2}.$$
I proved this with Muirhead's inequality, but: is there a better/more elegant way?
Given that $xyz=1$, prove that $\frac{x}{1+x^4}+\frac{y}{1+y^4}+\frac{z}{1+z^4}\le \frac{3}{2}$
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multivariable-calculus
inequality
contest-math
symmetric-functions
tangent-line-method
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0You must include your own work while posting these questions. This helps others to answer your questions better. – 2017-01-03
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0I know that but i don't know how to start. – 2017-01-03
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0@Deni Did you try to multiply both sides with $(1+x^4)(1+y^4)(1+z^4)$, do the manipulations and then use known inequalities, such the AM-GM? I know this method is not elegant, but trust me, sometimes most people can't find other way to prove some inequalities, especially the ones which state that a sum of fractions is less than an other quantity, like your inequality. – 2017-01-03
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0@SachpazisStelios I didn't, but i'll try it now. Thanks for the ideea. I'll let to know if it works. – 2017-01-03
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0@Deni Nice inequality! Where did you get this problem? – 2017-01-03
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0@MichaelRozenberg Very beautiful demonstration. Well, i found it in a very old book of contest problems. I have more like this .. Thanks for the answer. :) – 2017-01-03
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0@MichaelRozenberg I can tell you another if you want. I proved this inequality with Muirhead, but this is more elegant. – 2017-01-03
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0In general, whenever you prove an inequality with Muirhead, this implies that it can also be solved with one or more applications of AM-GM. – 2018-09-07
1 Answers
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Easy to see that it's enough to prove our inequality for positive variables.
We need to prove that $\sum\limits_{cyc}\left(\frac{1}{2}-\frac{x}{1+x^4}\right)\geq0$ or $\sum\limits_{cyc}\left(\frac{1}{2}-\frac{x}{1+x^4}-\frac{1}{2}\ln{x}\right)\geq0$.
Let $f(x)=\frac{1}{2}-\frac{x}{1+x^4}-\frac{1}{2}\ln{x}$.
We see that $f'(x)=\frac{3x^4-1}{(1+x^4)^2}-\frac{1}{2x}=\frac{(1-x)(x^7+x^6+x^5-5x^4-3x^3-3x^2-3x-1)}{2x(1+x^4)^2}$,
which says that $x_{min}=1$ and for $x=x_1$, where $x_1=1.64...$, $f'(x_1)=0$
and $f$ is a decreasing function on $[x_1,+\infty)$
and since $\lim\limits_{x\rightarrow+\infty}f(x)=-\infty$, there is an unique $x_2\in[x_1,+\infty)$, for which $f(x_2)=0$.
Indeed, $x_2=2.335...$ and $f(1)=0$,
which says that for $\max\{x,y,z\}<2$ our inequality is proven.
Let $x\geq2$ and $g(x)=\frac{x}{1+x^4}$.
$g'(x)=\frac{1-3x^4}{(1+x^4)^2}$, which says that $g$ is a decreasing function on $[2,+\infty)$
and $\sum\limits_{cyc}g(x)\leq2g\left(\frac{1}{\sqrt[4]3}\right)+g(2)<\frac{3}{2}$.
Done!
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1For your last step, what yo are saying is to take $2g_{\max} + g(2)$, and that $g_{\max}$ happens at $x=1/\sqrt[4]{3}$. You accidentally wrote $x=1/\sqrt[3]{3}$, but the relationship still holds at the actual max, which gives 1.257 instead of 1.244. – 2017-03-21
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0@Mark Fischler I fixed my post. Thank you very much dear friend! – 2017-03-21