1. ## Inequality proof

Prove [1/2(a+b)]^2 < or equal to 1/2(a^2+b^2). where a, b are real numbers.

Can't figure out how to end up with a contradiction or if thats even how to approach the problem.

2. It is well known that $\displaystyle a^2 + b^2 \geqslant 2ab$.
It is key to doing this problem.
$\displaystyle \left[ {\frac{{a + b}}{2}} \right]^2 = \frac{{a^2 + {\color{blue}2ab} + b^2 }}{4} \leqslant \frac{{2a^2 + 2b^2 }}{4} = \frac{{a^2 + b^2 }}{2}$

3. Is that some kind of theorem though, our professor asks us to be very specific. We cannot assume anything unless it was proven in our book or we prove it oursleves.

4. It is well known that $\displaystyle a^2 + b^2 \geqslant 2ab$.
PROOF:
$\displaystyle \begin{array}{rcl} {\left( {a - b} \right)^2 } & \geqslant & 0 \\ {a^2 - 2ab + b^2 } & \geqslant & 0 \\ {a^2 + b^2 } & \geqslant & {2ab} \\ \end{array}$

5. wow that makes a lot of sense...thanks!