<s>[INST] How find the maximum of $\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{z}{1+z^2}$ let $x,y,z$ are postive numbers,and such
$$xy+zx+yz=1$$
find the maxum of
$$\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}+\dfrac{z}{1+z^2}$$
my try
note
$$x^2+1=x^2+xy+yz+xz=(x+y)(x+z)$$
$$y^2+1=(y+x)(y+z)$$
$$z^2+1=(z+x)(z+y)$$
$$\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}+\dfrac{z}{1+z^2}=\dfrac{y+z+x+z+z(x+y)}{(x+y)(y+z)(x+z)}$$
I think we must use
$$8(a+b+c)(ab+bc+ac)\le  9(a+b)(b+c)(a+c)$$
$$LHS\le \dfrac{9}{8}\dfrac{x+y+z+z(x+y+1)}{x+y+z}=\dfrac{9}{8}(1+\dfrac{z(x+y+1)}{x+y+z})$$
then follow I can't works,Thank you 
 [/INST] Due to symmetry of the formula, you can suppose x=y=t. Hence
 $$t^2+2tz=1$$
 $$z=\frac{1-t^2}{2t}$$
$$f(t)=\frac{2}{1+t^2}+\frac{2t(1-t^2)}{4t^2+(1-t^2)^2}=2\frac{1+t+t^2-t^3}{(1+t^2)^2}$$
$$f'(t)=2\frac{(-3t^2+2t+1)(1+t^2)-4t(-t^3+t^2+t+1)}{(1+t^2)^3}$$
$$f'(t)=2\frac{t^4-2t^3-6t^2-2t+1}{(1+t^2)^3}$$
$P(t)=t^4-2t^3-6t^2-2t+1=(t+1)(t^3-3t^2-3t+1)=(t+1)^2(t^2-4t+1)$
Hence there are extremum in $t=-1$ (not really as it is a double root), and $t={2\pm\sqrt{3}}$. You can easily verify the maximum is for $2-\sqrt{3}$.
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