<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] Let $a = \sqrt{2} \cos u, b = \sqrt{2} \sin u$ which immediately satisfies the condition. Using Simon's Favorite Factoring Trick, $9 + 3a + 3b + ab = (a + 3)(b + 3)$, so by C-S:
$$(\sqrt{2} \cos u + 3)(\sqrt{2} \sin u + 3) ≥(\sqrt{\sqrt{2} \cos u} \sqrt{\sqrt{2} \sin u} + \sqrt{3} \sqrt{3})^2$$
$$= (\sqrt{2 \cos u \sin u} + 3)^2 = (\sqrt{\sin(2u)} + 3)^2$$
There are two values which have $\sqrt{2} \cos u$, $\sqrt{2} \sin u$ as their square. Choosing the negative value for the minimum, this is greater or equal to $(-1 + 3)^2 = 4$, hence $3a + 3b + ab ≥ 4 - 9 = -5$.
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