Prove that $\sqrt{2a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4$ when $a+b=2$.

Question

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?

Answer 1

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 $$...\geq ({3\over 4}a+{5\over 4})+ ({3\over 4}b+{5\over 4})=4$$

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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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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 $a+b=2$.

Answer 6

$2a^2+b+2=2a^2+3-a=(\frac{a^2+1}{2}-a)+(\frac{3a^2+5}{2})\geq\frac{3a^2+5}{2}$.

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 $\sqrt{2a^2+b+2}+\sqrt{2b^2+a+2}\geq\frac{(3a+5)+(3b+5)}{4}=\frac{3(a+b)+10}{4}=\frac{16}{4}=4$