<s>[INST] How find this maximum $\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{1+c^2}$ let $a,b,c>0$ ,and such $a+b+c=3$.prove that
$$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{1+c^2}<\dfrac{11}{5}$$
if this problem  find this minimum,then 
$$f(x)=\dfrac{1}{x^2+1}\ge ax+b$$
where $a=f'(1),a+b=f(1)$
since $$f'(x)=-\dfrac{2x}{(1+x^2)^2}\Longrightarrow f'(1)=-\dfrac{1}{2}$$
so
$$a=-\dfrac{1}{2},b=1$$
and
$$\dfrac{1}{1+x^2}+\dfrac{1}{2}x-1=\dfrac{x(x-1)^2}{(x^2+1)^2}\ge 0$$
then
$$\dfrac{1}{1+x^2}\ge-\dfrac{1}{2}x+1$$
so
$$\sum_{cyc}\dfrac{1}{1+a^2}\ge\sum_{cyc}-\dfrac{1}{2}a+1=-\dfrac{1}{2}(a+b+c)+3=\dfrac{3}{2}$$
But for maximum,I can't prove it.Thank you
 [/INST] First show that $\dfrac{1}{1+x^2} < \dfrac{63 - 19x}{60}$ for $0 < x < 3$.
Then $\text{LHS} < \dfrac{3\cdot63 - 19(a + b + c)}{60} = \dfrac{189 - 57}{60} = \text{RHS}$.
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