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?

Since: $$\sqrt{2 a^2+b+1}= \sqrt{2 a^2-a+3}\geq {3\over 4}a+{5\over 4}$$
and the same for the other root, we have $$...\geq ({3\over 4}a+{5\over 4})+
({3\over 4}b+{5\over 4})=4$$

Here is a detailed proof for $$\sqrt{2 a^2-a+3}\geq {3\over 4}a+{5\over 4}$$
It is equivalent to $$16(2a^2-a+3)\geq (3a+5)^2$$
and this is equivalent to $$23a^2-46a+23\geq 0$$
or $$23(a-1)^2\geq 0$$
which is obviously true.
