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