# Given that $xyz=1$, prove that $\frac{x}{1+x^4}+\frac{y}{1+y^4}+\frac{z}{1+z^4}\le \frac{3}{2}$

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?

#### Solutions Collecting From Web of "Given that $xyz=1$, prove that $\frac{x}{1+x^4}+\frac{y}{1+y^4}+\frac{z}{1+z^4}\le \frac{3}{2}$"

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!