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 $x=a+1,y=b+1$. Then the statement is equivalent to $$(x-1)^2+(y-1)^2=2\Rightarrow 3(x-1)+3(y-1)+(x-1)(y-1)\geq -5,$$ or $$x^2+y^2=2x+2y\Rightarrow (2x+2y)+xy\geq 0,$$ which is clear, since $$(x^2+y^2)+xy\geq 0.$$
