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?
 A: (This answer uses a bit of elementary calculus, which could be circumvented with some epsilonology. It serves as a pedestrian introduction to math110's answer.) We have a linear constraint on the variables; and the target function is a linear combination of the same function applied to each variable: namely, the function given in cartesian form by
$$y=\dfrac{1}{11+x^2}.$$This is a bell-shaped curve and, to exploit the linear structures already noticed, we seek a linear upper approximant for it, over the interval $[0,1],$ which is exact at the critical point $(x,y)=(1,1/12).$ The gradient of the curve at this point is $-2\times 1/(11+1^2)^2=-1/72.$ A little coordinate geometry gives the equation of the tangent at $x=1$ as$$y=\dfrac{7-x}{72}.$$That this is an upper approximant for the curve over $[0,1]$ is shown rigorously in math110's answer, and the rest of the deduction can be followed in that answer.
A: 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
$$\dfrac{1}{11+a^2}+\dfrac{1}{11+b^2}+\dfrac{1}{11+c^2}+\dfrac{1}{11+d^2}\le\dfrac{1}{72}(7+7+7+7-(a+b+c+d))=\dfrac{1}{3}$$
A: 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 :
$$\frac{a^2}{b+c+d}\ge\frac{b^2}{a+c+d}\ge \frac{c^2}{b+a+d}\ge\frac{d^2}{b+c+a}\\\Longrightarrow\dfrac{1}{11+a^2}+\dfrac{1}{11+b^2}+\dfrac{1}{11+c^2}+\dfrac{1}{11+d^2} \leq \\\dfrac{1}{11+a^2}+\dfrac{1}{11+a^2\cdot\frac{(4-b)^2}{(4-a)^2}}+\dfrac{1}{11+a^2\cdot\frac{(4-c)^2}{(4-a)^2}}+\dfrac{1}{11+a^2\cdot\frac{(4-d)^2}{(4-a)^2}} \leq \frac{4}{11+a^2}\le\frac{4}{12}=\frac{1}{3}$$
A: 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 !
