Let $a,b,c,d\in \mathbb{R}$ and $a+b+c+d=2$. Prove that $$\frac{a}{a^2-a+1}+\frac{b}{b^2-b+1}+\frac{c}{c^2-c+1}+\frac{d}{d^2-d+1}\le \frac{8}{3}.$$
We have $$\frac{a}{a^2-a+1}\le \frac{4}{3}a\Longleftrightarrow \frac{-a\left(2a-1\right)^2}{3\left(a^2-a+1\right)}\le 0,$$ (Right)
So $$LHS\le \frac{4}{3}\left(a+b+c+d\right)=\frac{8}{3}=RHS$$
When $a=b=c=d=\frac{1}{2}$
But also $a=b=c=1;d=-1$ i don't know how to solve it.
First, we simplify a bit by transforming $x = a-\frac12, y = b - \frac12, z = c - \frac12, w = d - \frac12$, so that we need to prove equivalently with $x+y+z+w=0$, the inequality $$\sum \frac{2x+1}{4x^2+3} \leqslant \frac43$$
Here $\sum$ represents cyclic sums. Also as $\displaystyle 1 - 2\cdot\frac{2x+1}{4x^2+3} = \frac{(2x-1)^2}{4x^2+3}$, we may equivalently show $$\sum \frac{(2x-1)^2}{4x^2+3} \geqslant \frac43$$
Now we need an upper bound on the denominator; for this, note
$$4x^2+3=3x^2+(y+z+w)^2+3\leqslant 3x^2+3(y^2+z^2+w^2)+3 = 3\sum x^2+3$$ Thus we have a common denominator now, therefore, $$\sum \frac{(2x-1)^2}{4x^2+3} \geqslant \frac{\sum (2x-1)^2}{3\sum x^2+3} = \frac{4\sum x^2-4\sum x+4}{3(\sum x^2+1)}=\frac43$$