Beckenbach Introduction to Inequalities Chapter 2: Show $(a+b)/2 \le ( (a^2 + b^2 )/2)^{1/2}$

Question

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

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

Thanks

Answer 1

1. 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.

2. 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.

Answer 2

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 $(a-b)^2\geq 0$.

Answer 3

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 $(a+b)^2$