<s>[INST] Prove that $\sqrt{2a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4$ when $a+b=2$. Suppose that $a,b$ are non-negative numbers such that $a+b=2$. I want to prove $$\sqrt{2 a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4.$$
My attempt: I tried proving $$2a^2+b+1\geq 4$$ but this seems wrong (try $a=0,b=1$).
How can I prove the above result?
 [/INST] With $$b=2-a$$ we have to prove:
$$\sqrt{4a^2-a+3}\geq 4-\sqrt{2a^2-7a+9}$$ if $$4-\sqrt{2a^2-7a+9}\le 0$$ we have nothing to prove, in the other case we get
$$-3a+11\le 4\sqrt{2a^2-7a+9}$$ If $$-3a+11\le 0$$ then is all clear, in the other case we can square and collecting like terms we get
$$-21(a-1)^2\le 0$$ and this is true.
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