If $a+b+c+d = 2$, then $\frac{a^2}{(a^2+1)^2}+\frac{b^2}{(b^2+1)^2}+\frac{c^2}{(c^2+1)^2}+\frac{d^2}{(d^2+1)^2}\le \frac{16}{25}$ If  $a+b+c+d = 2$, prove that
$$\dfrac{a^2}{(a^2+1)^2}+\dfrac{b^2}{(b^2+1)^2}+\dfrac{c^2}{(c^2+1)^2}+\dfrac{d^2}{(d^2+1)^2}\le \dfrac{16}{25}$$
Also $a,b,c,d \ge 0$.
 A: Assume $0 \le a \le b \le c \le d$, with $a+b+c+d=2$,
Then $(48a-4)(a^2+1)^2-125a^2 = (2a-1)^2(12a^3+11a^2+32a - 4) \ge 0$, for $a \ge \frac{1}{8}$;  
(See that $32a-4 \ge 0$ and $12a^3+11a^2$ is positive). 
That is $\dfrac{a^2}{(a^2+1)^2} \le \dfrac{48a-4}{125}$, and similarly for $b,c,d$ and adding them we have $\sum\limits_{a,b,c,d} \dfrac{a^2}{(a^2+1)^2} \le \sum\limits_{a,b,c,d} \dfrac{48a-4}{125}=\dfrac{80}{125}=\dfrac{16}{25}$.
The case $a < \frac{1}{8}$, 
We have $(540x + 108)(x^2+1)^2 - 2197x^2 = (3x-2)^2(60x^3+92x^2+216x+27) \ge 0$, for $x \ge 0$.
Thus $\sum\limits_{x \in \{b,c,d\}} \dfrac{x^2}{(x^2+1)^2} \le \sum\limits_{x \in \{b,c,d\}} \dfrac{540x+108}{2197}=\dfrac{108}{169}-\dfrac{540a}{2197}$
and, $\dfrac{a^2}{(a^2+1)^2} < a^2 < \dfrac{a}{8} < \dfrac{540a}{2197}$.
So, $\sum\limits_{a,b,c,d} \dfrac{a^2}{(a^2+1)^2} \le \dfrac{108}{169} < \dfrac{16}{25}$
Equality occurs iff $a=b=c=d=\dfrac{1}{2}$.
A: Let us consider the function 
$$
\frac{x^2}{(1+x^2)^2} +
\frac{y^2}{(1+y^2)^2} +
\frac{z^2}{(1+z^2)^2} +
\frac{t^2}{(1+t^2)^2} +
\lambda(x+y+z+t-2)
$$and write the first order conditions:
$$
0 = -\frac{2x(x^2-1)}{(1+x^2)^3} + \lambda,
$$and the same equation for $y,z,t$. In particular,
$$
\frac{x(x^2-1)}{(1+x^2)^3} = \frac{y(y^2-1)}{(1+y^2)^3}
= \frac{z(z^2-1)}{(1+z^2)^3} = \frac{t(t^2-1)}{(1+t^2)^3}
$$
Taking a look at the variations of this function,

we see that if $x\neq y$ then $x,y < 1$ and $x,y,z,t\in \{u,v\}$, with
$$
\begin{cases}
\frac{u(u^2-1)}{(1+u^2)^3} = \frac{v(v^2-1)}{(1+v^2)^3} \\
u+v=1\\
u<v.
\end{cases}
$$
Then an analysis  proves that 
$(u,v) = (0,1)$:

Eventually, compute the two potential extrema:
$$
\frac{1^2}{(1+1^2)^2} + \frac{1^2}{(1+1^2)^2}+0+0 =\frac 12;\\
\frac{(1/2)^2}{(1+(1/2)^2)^2} + \frac{(1/2)^2}{(1+(1/2)^2)^2}+
\frac{(1/2)^2}{(1+(1/2)^2)^2} + \frac{(1/2)^2}{(1+(1/2)^2)^2} = \frac {16}{25}.
$$
