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 
 A: Trigonometric substitution seems easier here.  We may set $x = \tan\frac{A}{2}, y = \tan\frac{B}{2}, z = \tan\frac{C}{2}$, where $A, B, C$ are angles of a triangle, so that the constraint is satisfied.  Then we have to find the maximum of 
$$f = \dfrac{1}{1+\tan^2\frac{A}{2}}+\dfrac{1}{1+\tan^2\frac{B}{2}}+\dfrac{\tan\frac{C}{2}}{1+\tan^2\frac{C}{2}} = \cos^2\dfrac{A}{2} + \cos^2\dfrac{B}{2}+ \dfrac{\sin C}{2}$$
Using $\cos \theta = 2 \cos^2 \frac{\theta}{2} - 1$, 
$$f = \dfrac{\cos A + 1}{2} + \dfrac{\cos B + 1}{2}+ \dfrac{\sin C }{2} = 1+\dfrac{\cos A + \cos B + \sin C}{2}$$
So we need to find the maximimum of $\cos A + \cos B + \sin C$, where $A, B, C$ are angles of a triangle. Now,
$$\cos A + \cos B + \sin C = \cos A + \cos B + \sin (\pi - (A+B)) \\
= \frac{2}{\sqrt3}\left(\frac{\sqrt3}{2} \cos A + \frac{\sqrt3}{2} \cos B\right) + \frac{1}{\sqrt3}\left(\sqrt3 \sin A \cos B + \sqrt 3 \cos A \sin B\right)\\
\le \frac{1}{\sqrt3}\left(\frac{3}{4} + \cos^2 A + \frac{3}{4}+ \cos^2 B\right) + \frac{1}{2\sqrt3}\left(3 \sin^2 A +\cos^2 B +  \cos^2 A +3\sin^2 B\right)\\
= \frac{\sqrt3}{2}+\frac{\sqrt3}{2}(\sin^2 A + \cos^2A) + \frac{\sqrt3}{2}(\sin^2 B + \cos^2 B) = \frac{3\sqrt 3}{2}$$
Hence $f \le 1 + \frac{3\sqrt3}{4}$, and equality is obtained when the AM-GMs we used have equality, i.e. $A = B = \frac{\pi}{6}, C = \frac{2\pi}{3}$ or $x = y = 2 - \sqrt{3}, z = \sqrt{3}$
A: 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 $2-\sqrt{3}$.
