For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$. For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$.
My attempt: We have
$$2(a^2+b^2)\geq (a+b)^2$$
so $$-2\leq a+b \leq 2$$
In other hand $$ab=\frac{(a+b)^2-2}{2}=(a+b)^2-1$$

let $a=x-y$ and $b=x+y$
$a^2+b^2=2 \implies x^2+y^2=1 $
substituting $y$ given that $x^2+y^2=1 \implies $
the problem now is to prove that $6x+x^2-y^2 \ge -5 \iff 2x^2+6x-1 \ge -5 $ (1)
$y^2=1-x^2 \ge 0 \implies -1 \ge x \ge 1$
it's enough to see that (1) $\iff (x-1)(2x-4) $
which is easy to check that it's positive when $-1 \ge x \ge 1$
