Answer
The quadratic mean is greater or equal to the arithmetic mean
Work Step by Step
Let $x$ and $y$ be two real numbers.
The arithmetic mean is:
$M_a=\dfrac{x+y}{2}$,
while the quadratic mean is:
$M_q=\sqrt{\dfrac{x^2+y^2}{2}}$
We have:
$(x-y)^2\geq 0$
$x^2-2xy+y^2\geq 0$
$x^2+y^2\geq 2xy$
$x^2+y^2+x^2+y^2\geq 2xy+x^2+y^2$
$2(x^2+y^2)\geq (x+y)^2$
$\dfrac{2(x^2+y^2)}{4}\geq \dfrac{(x+y)^2}{4}$
$\dfrac{x^2+y^2}{2}\geq \left(\dfrac{x+y}{2}\right)^2$
$\sqrt{\dfrac{x^2+y^2}{2}}\geq \dfrac{x+y}{2}$
$M_q\geq M_a$
Therefore the quadratic mean is greater or equal to the arithmetic mean.
