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}

More math olympiad problems later,

Nalin Pithwa

 

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s