Just take the average, right?

There are ways to think this through intuitively, but here is an algebraic proof:

$\begin{aligned} x&=\frac{a^2+b^2}{2} \\ \\ &=\frac{a^2+2ab+b^2}{4}+\frac{a^2-2ab+b^2}{4} \\ \\ &=\left(\frac{a+b}{2}\right)^2+\left(\frac{a-b}{2}\right)^2 \\ \\ \sqrt{x}&=\sqrt{\left(\frac{a+b}{2}\right)^2+\left(\frac{a-b}{2}\right)^2} \end{aligned}$

Since $\left(\dfrac{a-b}{2}\right)^2$ is a positive number, $\sqrt{x}>\dfrac{a+b}{2}$ .

A small side note: I wrote this problem to demonstrate an interesting fact about estimating square roots. When you do a linear approximation of a square root, which is what the expression $\frac{a+b}{2}$ is doing for $\sqrt{x}$ , your approximation tends to be less than the actual value.

By Cauchy-Schwarz Inequality,

$2(a^2+b^2)\geq(a+b)^2$

Since $a$ and $b$ are distinct, we remove the equality case. The result follows.

$x>\dfrac{(a+b)^2}{4}$

$\boxed{\sqrt{x}>\dfrac{a+b}{2}}$

Let $x = \frac {a^2 + b^2}{2}$ than $\sqrt{x} = \sqrt{\frac {a^2 + b^2}{2}}$ . you can see from here that $\sqrt{x}$ denotes the Quadratic mean of a and b. by QM-AM-GM-HM inequality: $\sqrt{x}$ $\geq$ $\frac{a+b}{2}$ , but you mentiond that a and b are distinct, therefore the equality sign drops and we are left with : $\sqrt{x}$ > $\frac{a+b}{2}$ . (*NOTE: I am sorry for my bad Latex skills, i am a beginner at it, and if I have a mistake in my solution, please tell me so I would fix it.)

The given question compares the average value, and the RMS (root mean square value). RMS>Avg.