# Proving an Inequality

• Sep 4th 2008, 10:32 AM
hockey777
Proving an Inequality
So i did this last week, and nailed it. The teacher hasn't given it back yet and I was trying to do it again, now I'm stuck.

I need to prove that

$[\dfrac{1}{2}(a+b)] ^2\leq \dfrac{1}{2}(a^2+b^2)$

I do know that the inequality is equivalent to $0 \leq a^2 -2ab +b^2$
• Sep 4th 2008, 10:45 AM
flyingsquirrel
Hello,
Quote:

Originally Posted by hockey777
So i did this last week, and nailed it. The teacher hasn't given it back yet and I was trying to do it again, now I'm stuck.

I need to prove that

$[\dfrac{1}{2}(a+b)] ^2\leq \dfrac{1}{2}(a^2+b^2)$

I do know that the inequality is equivalent to $0 \leq a^2 -2ab +b^2$

And you also know that $a^2-2ab+b^2=(a-b)^2$ and that the square of a real number is non negative, don't you ?
• Sep 4th 2008, 12:43 PM
hockey777
Got it, sorry. I was trying to do it a weird way with theorems.
• Sep 4th 2008, 12:58 PM
Plato
It is well known that $a^2 + b^2 \ge 2ab$.
Now add to both parts: $2a^2 + 2b^2 \ge a^2 + 2ab + b^2 = \left( {a + b} \right)^2$.
Regroup: $\frac{1}{2}\left( {a^2 + b^2 } \right) \ge \frac{1}{4}\left( {a + b} \right)^2 = \left[ {\frac{1}{2}\left( {a + b} \right)} \right]^2$