<s>[INST] 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 $a,b,c,d \ge 0$.
 [/INST] 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 $a=b=c=d=\dfrac{1}{2}$.
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