<s>[INST] Beckenbach Introduction to Inequalities Chapter 2: Show $(a+b)/2 \le ( (a^2 + b^2 )/2)^{1/2}$ I'm having trouble understanding the following problem
Problem

Beckenbach, Chapter 2 Pg 24 Ex 1
  $$
\text{Show the following for all a, b}\quad
\frac{(a+b)}{2} \le \left(\frac{a^2 + b^2}{2}\right)^\frac{1}{2}
$$

The book provides answers in the back, the answer is shown as
$$
\text{equivalent to}\quad(a - b)^2 \ge 0
$$
My attempt 
Thanks to this answer: Proving the inequality $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$ I was able to show the following
$$
\begin{align}
\frac{a+b}{2} &\le \sqrt{\frac{a^2 + b^2}{2}} \\
\frac{a+b}{2} &\le \sqrt{\frac{(a+b)^2 + (a-b)^2}{4}}\\
\frac{a+b}{2} &\le \sqrt{\left(\frac{a+b}{2}\right)^2\left(1 + \left(\frac{a-b}{a+b}\right)^2\right)}\\
\frac{a+b}{2} &\le \frac{a+b}{2}\sqrt{1 + \left(\frac{a-b}{a+b}\right)^2}\\
0 &\le \sqrt{1 + \left(\frac{a-b}{a+b}\right)^2} - 1 \\
\text{Since}\quad\left(\frac{a-b}{a+b}\right)^2 \ge 0\quad\text{we're done}
\end{align}
$$
This feels ugly and I don't think is the way Beckenbach intended us to solve, considering this is in Chapter 2.
Question

  
*
  
*Is my attempt valid?
  
*Is there a more elegant way to show this, using $(a-b)^2 \ge 0$?
  

Thanks
 [/INST] Squaring both sides...
$\cfrac{(a+b)^2}{4}\le \cfrac{a^2+b^2}{2}$
$\cfrac{a^2+2ab+b^2}{4}\le \cfrac{a^2+b^2}{2}$
$2a^2+4ab+2b^2\le4(a^2+b^2)$
$2a^2+4ab+2b^2\le4[(a+b)^2-2ab]$
$(a+b)^2\le2(a+b)^2-4ab$
Which proves your inequality, since any numbers $0<x<1$ will produce a positive result, as $-4ab$ will equal to a value smaller than $(a+b)^2$
</s>