Let $a;b;c \in \mathbb{R} $ such that $a+b+c=0$ Prove that: $P=\dfrac{a-1}{a^2+8}+\dfrac{b-1}{b^2+8}+\dfrac{c-1}{c^2+8} \geq -\dfrac{3}{8}$
I tried to do this: $\dfrac{8a-8}{a^2+8}+2+\dfrac{8b-8}{b^2+8}+2+\dfrac{8c-8}{c^2+8}+2 \geq 3$
$\Leftrightarrow \dfrac{(a+2)^2}{a^2+8}+\dfrac{(b+2)^2}{b^2+8}+\dfrac{(c+2)^2}{c^2+8} \geq \dfrac{3}{2}$
Then WLOG $c=max\{a;b;c\} \Rightarrow c \geq 0$
I hope I can prove: $\dfrac{(a+2)^2}{a^2+8}+\dfrac{(b+2)^2}{b^2+8} \geq f(c)$
So I only need to find minimum of $f(c)+\dfrac{(c+2)^2}{c^2+8}$
But I can't find $f(c)$, I stuck here.
Your idea is very good and we will use that :)
As $a+b+c=0$ we can find a pair such that they have the same sign hence WLOG $ab\ge 0$. Now from your attempt we have to prove $$\Leftrightarrow \dfrac{(a+2)^2}{a^2+8}+\dfrac{(b+2)^2}{b^2+8}+\dfrac{(c+2)^2}{c^2+8} \geq \dfrac{3}{2}$$ by C-S $$\dfrac{(a+2)^2}{a^2+8}+\dfrac{(b+2)^2}{b^2+8}\ge \frac{{(a+b+4)}^2}{a^2+b^2+16}=\frac{{(4-c)}^2}{a^2+b^2+16}\ge \frac{{(4-c)}^2}{c^2+16} $$ here we used $a^2+b^2\le {(a+b)}^2=c^2\iff 2ab\ge 0$
it remains to prove $$\frac{{(4-c)}^2}{c^2+16}+\frac{{(c+2)}^2}{c^2+8}\ge \frac{3}{2}$$ $$\iff \frac{c^2{(c-4)}^2}{2(c^2+16)(c^2+8)}\ge 0$$ which is true
Equality when $a=b=c=0$,
$c=4,a=-2,b=-2$(upto permutations)