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?
 A: We may set $a=1-t, b=1+t$ and study the behaviour of 
$$ f(t)=\sqrt{2(1-t)^2+(1+t)+1}+\sqrt{2(1+t)^2+(1-t)+1}=\sqrt{4-3t+t^2}+\sqrt{4+3t+t^2} $$
over $[-1,1]$. $f(t)$ is even and convex (as the sum of two convex functions), hence it attains its minimum value at the origin, QED.
A: By Minkowski (triangle inequality) we obtain:
$$\sum_{cyc}\sqrt{2a^2+b+1}=\sum_{cyc}\sqrt{2a^2+\frac{b(a+b)}{2}+\frac{(a+b)^2}{4}}=$$
$$=\frac{1}{2}\sum_{cyc}\sqrt{9a^2+4ab+3b^2}=\frac{1}{2}\sum_{cyc}\sqrt{7a^2+b^2+8}\geq$$
$$\geq\frac{1}{2}\sqrt{7(a+b)^2+(b+a)^2+8(1+1)^2}=4.$$
We see that our inequality is true for all reals $a$ and $b$ such that $a+b=2$.
A: Suppose without loss of generality that $a\geqslant b$ so that $0\leqslant b \leqslant 1$ and write $a = 2- b$.
Our expression becomes
$$\sqrt{2(2-b)^2 + b + 1} + \sqrt{2b^2 - b + 3}$$
and this can be minimized for $b\in[0,1]$ with usual calculus techniques.
A: 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.
A: 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.
A: $2a^2+b+2=2a^2+3-a=(\frac{a^2+1}{2}-a)+(\frac{3a^2+5}{2})\geq\frac{3a^2+5}{2}$.
On the other hand by QM-AM $\sqrt{\frac{a^2+a^2+a^2+1+1+1+1+1}{8}}\geq\frac{3a+5}{8}$, therefore $\sqrt{3a^2+5}\geq\frac{3a+5}{\sqrt{8}}$. Now we return back: $\sqrt{2a^2+b+2}\geq\sqrt{\frac{3a^2+5}{2}}\geq\frac{3a+5}{\sqrt{2\times8}}=\frac{3a+5}{4}$. 
Therefore $\sqrt{2a^2+b+2}+\sqrt{2b^2+a+2}\geq\frac{(3a+5)+(3b+5)}{4}=\frac{3(a+b)+10}{4}=\frac{16}{4}=4$
