Mathematical Reflections 2012, Issue 4 - Problem J239

Problem:
Let $a$ and $b$ be real numbers so that $2a^2 + 3ab + 2b^2 \leq 7$. Prove that $$\max\{2a+b,2b+a\} \leq 4.$$

Proposed by Titu Andreescu.

Solution:

Suppose that $a \geq b$, so that $\max\{2a+b,2b+a\}=2a+b$. We argue by contradiction. Suppose that for some real numbers $a,b$ we have $2a+b>4$, i.e. $a>\dfrac{4-b}{2}$. Then
$$7 \geq 2a^2 + 3ab + 2b^2 > b^2+2b+8,$$ which gives $(b+1)^2 < 0$, contradiction.