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 $a,b,c,d \ge 0$.

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:
$$
\frac{1^2}{(1+1^2)^2} + \frac{1^2}{(1+1^2)^2}+0+0 =\frac 12;\\
\frac{(1/2)^2}{(1+(1/2)^2)^2} + \frac{(1/2)^2}{(1+(1/2)^2)^2}+
\frac{(1/2)^2}{(1+(1/2)^2)^2} + \frac{(1/2)^2}{(1+(1/2)^2)^2} = \frac {16}{25}.
$$
