An inequality for Mathematics Olympiads

Prove that for any real numbers a and b, both greater than zero, we have that:

$\frac{(a+b)^{2}}{2}+\frac{a+b}{4} \geq a \sqrt{b} + b \sqrt{a}$

All-Soviet Union Olympiad, 1984.

Proof:

It does require some imagination to solve this problem. Firstly, note that it seems similar to the inequality — $AM \geq GM \geq HM$ .

Here is the solution:

$2a^{2}+ \frac{b}{2} \geq 2a \sqrt{b}$

$2b^{2}+\frac{a}{2} \geq 2b \sqrt{a}$

$2ab+\frac{b}{2} \geq 2b \sqrt{a}$

$2ab + \frac{a}{2} \geq 2a \sqrt{b}$

By adding all these, we get

$2(a+b)^{2}+a+b \geq 4a \sqrt{b} + 4b \sqrt{a} \Longrightarrow \frac{(a+b)^{2}}{2}+\frac{a+b}{4} \geq a \sqrt{b} + b \sqrt{a}$

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