<s>[INST] 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?
 [/INST] since
$$\dfrac{1}{11+x^2}\le\dfrac{7-x}{72}$$
because
$$\Longleftrightarrow  (7-x)(11+x^2)\ge 72$$
$$\Longleftrightarrow x^3-7x^2+11x-5\le0 $$
$$\Longleftrightarrow (x-1)^2(x-5)\le 0$$
this is true
so
$$\dfrac{1}{11+a^2}+\dfrac{1}{11+b^2}+\dfrac{1}{11+c^2}+\dfrac{1}{11+d^2}\le\dfrac{1}{72}(7+7+7+7-(a+b+c+d))=\dfrac{1}{3}$$
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