If $a,b,c,d>0$ and $a+b+c+d=4$, then $\frac{1}{11+a^2}+\frac{1}{11+b^2}+\frac{1}{11+c^2}+\frac{1}{11+d^2} \leq \frac {1}{3}$ Prove if $a,b,c,d>0$ and $a+b+c+d=4$, then $$\dfrac{1}{11+a^2}+\dfrac{1}{11+b^2}+\dfrac{1}{11+c^2}+\dfrac{1}{11+d^2} \leq \dfrac {1}{3}$$
This was an Inequality Olympiad Problem1.
I  proved  by  using  Lagrange  Multipliers  method.  Can  you  do  without  calculus?

oh sorry, there is some detail in my post ! i omit it and i write it here :
$\Longrightarrow\dfrac{1}{11+a^2}+\dfrac{1}{11+b^2}+\dfrac{1}{11+c^2}+\dfrac{1}{11+d^2} \leq \\\dfrac{1}{11+d^2}+\dfrac{1}{11+d^2\cdot\frac{(4-b)^2}{(4-d)^2}}+\dfrac{1}{11+d^2\cdot\frac{(4-c)^2}{(4-d)^2}}+\dfrac{1}{11+d^2\cdot\frac{(4-a)^2}{(4-d)^2}} $
this inequality holds, that is because if we assume :
$a=\frac{1}{x_{1}}$$,b=\frac{1}{x_{2}}$$,c=\frac{1}{x_{3}}$$,d=\frac{1}{x_{4}}$
then, our inequality transform to :
$\Longrightarrow\dfrac{x_{1}^2}{11x_{1}^2+1}+\dfrac{x_{2}^2}{11x_{2}^2+1}+\dfrac{x_{3}^2}{11x_{3}^2+1}+\dfrac{x_{4}^2}{11x_{4}^2+1}\le{\dfrac{4x_{4}^2}{11x_{4}^2+1}}\le\frac{1}{3}$
and $f(x)=\dfrac{x^2}{11x^2+1}$ monotone increase in its domain !
the hint is :
your inequality reach its upper bound as :
$a=b=c=d$ and $d$ can be replaced by $a$ !
so our inequality seems feasible !
