<s>[INST] 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 $$ab=\frac{(a+b)^2-2}{2}=(a+b)^2-1$$
 [/INST] Observe,
$$a^2+b^2=2\implies (a+b)^2=2(ab+1)$$
It remains to show that
$$3(a+b)+(ab+1)+4\ge0\;\Longleftrightarrow\;3(a+b)+\frac{(a+b)^2}{2}+4\ge0$$
Let $a+b=x$,  we have,
$$3x+\frac{x^2}{2}+4\ge0\;\Longleftrightarrow\;x^2+6x+8\ge0\;\Longleftrightarrow\;(x+2)(x+4)\ge0$$
$$\therefore\;x\le-4 \;\;\text{or}\;\;x\ge -2$$
Since you have shown that $a+b\ge-2$, the inequality must be true.

Alternate Solution:
Using AM-GM inequality,
$$\frac{a^2+b^2}{2}\ge\sqrt{a^2b^2}\implies 1\ge ab$$
Adding the inequalities,
$$3a+3b+ab\ge-5$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;1\ge ab$$
It remains to show that $a+b\ge-2$ which we know is true.
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