How can I prove that $y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}>0$ when $x>0$ and $1I would like to prove that
$$y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}>0$$
for all real numbers $x > 0$ and $1 < y < 1.5$. This seems true when plotted on WolframAlpha, but I don't know how to prove it. I tried replacing some of the terms using the given inequalities to obtain a simpler function, but any perturbation I make seems to render the inequality untrue. How would you approach this problem?
 A: Let $f(y)=y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}.$
Thus, $$f''(y)=-\frac{6xy^2}{(1+x^2)^2}+2x^3\left(\frac{y}{(1+y^2)^2}\right)'=$$
$$=-\frac{6xy^2}{(1+x^2)^2}+2x^3\left(\frac{1}{(1+y^2)^2}-\frac{4y^2}{(1+y^2)^3}\right)=$$
$$=-\frac{6xy^2}{(1+x^2)^2}+\frac{2x^3(1-3y^2)}{(1+y^2)^3}<0,$$
which says that $f$ is a concave function.
But the concave function gets a minimal value for extreme value of $y$,
which says that it's enough to prove our inequality for $y\in\{1,1.5\}.$
Can you end it now?
A: We have that
$$g(y)=\frac{\partial }{\partial y}\left(y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}\right) =1-\frac{2xy^3}{(1+x^2)^2}+\frac{2x^3y}{(1+y^2)^2}$$
with
$$g'(y)=\frac{\partial }{\partial y}\left(y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}\right) =-\frac{6xy^2}{(1+x^2)^2}-\frac{2x^3(-1+3y^2)}{(1+y^2)^3}<0$$
therefore the minimum is reached either for $y=1$ or $y=\frac32$, therefore we reduce to check and
$$x^5-\frac1{2}x^3-\frac{x}{2(x^2+1)^2}-x+1>0 \tag 1$$
$$x^5-\frac4{13}x^3-\frac{81}{32}\frac{x}{(x^2+1)^2}-x+\frac32>0 \tag 2$$
