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?

this is not a hard question !
firstly it is easy to see that :
$abcd\le1$
and we assume that :
$a\ge{b}\ge{c}\ge{d}$
then we use the inequality below :
$$\frac{a^2}{b+c+d}\ge\frac{b^2}{a+c+d}\ge \frac{c^2}{b+a+d}\ge\frac{d^2}{b+c+a}\\\Longrightarrow\dfrac{1}{11+a^2}+\dfrac{1}{11+b^2}+\dfrac{1}{11+c^2}+\dfrac{1}{11+d^2} \leq \\\dfrac{1}{11+a^2}+\dfrac{1}{11+a^2\cdot\frac{(4-b)^2}{(4-a)^2}}+\dfrac{1}{11+a^2\cdot\frac{(4-c)^2}{(4-a)^2}}+\dfrac{1}{11+a^2\cdot\frac{(4-d)^2}{(4-a)^2}} \leq \frac{4}{11+a^2}\le\frac{4}{12}=\frac{1}{3}$$
