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
 A: If left side is not positive then inequality obviously hold.
So assume it is nonegative. Square it: $${a^2+2ab+b^2\over 4}\leq {a^2+b^2\over 2}$$
Get rid of denumerators and you get
$$ a^2+2ab+b^2\leq 2a^2+2b^2$$
or $(a-b)^2\geq 0$.
A: *

*Your proof has a tiny issue you need to address (and then it'll be correct): you forgot absolute values around $\frac{a+b}{2}$ once you factor $\left(\frac{a+b}{2}\right)^2$ out of the square root.

*Note that
$$
\frac{a+b}{2} \leq \frac{\lvert a\rvert +\lvert b\rvert}{2}\tag{1}
$$
so it suffices to prove
$$
\frac{\lvert a\rvert +\lvert b\rvert}{2} \leq\left(\frac{a^2+b^2}{2}\right)^{1/2} \tag{2}
$$
Square both sides of (2): this is equivalent, this everything is non-negative: so (2) is equivalent to
$$
\frac{a^2+b^2+2\lvert a\rvert\lvert b\rvert}{4} \leq \frac{a^2+b^2}{2}
$$
in turn equivalent to
$$
a^2+b^2+2\lvert a\rvert\lvert b\rvert \leq 2a^2+2b^2
$$
in turn equivalent to
$$
0 \leq a^2+b^2+2\lvert a\rvert\lvert b\rvert \tag{3}
$$
and the RHS is equal to $(\lvert a\rvert -\lvert b\rvert )^2$.
So (3) holds.
A: 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 $(a+b)^2$
