Proving that either of the two numbers are greater or equal to their arithmetic mean.

Question

The arithmetic mean of two numbers $x$ and $y$ is $k$. Show that either $x \ge k$ or $y \ge k$. Can anyone explain what approach should I use?

Answer 1

This can be proved using the contrapositive of the given implication.

Suppose $x<k$ and $y<k$.

Then, the arithmetic mean, $a = \frac{x+y}{2}<\frac{2k}{2}=k$

That is, $a< k$.

Thus, if the arithmetic mean is $k$, then either $x\ge k$ or $y\ge k$.

Answer 2

Since $k = (x+y)/2$,

$\begin{array}\\ (x-k)(y-k) &=(x-(x+y)/2)(y-(x+y)/2)\\ &=\frac{x-y}{2}\frac{y-x}{2}\\ &=-\frac{(x-y)^2}{4}\\ &\le 0 \qquad\text{with equality only if }x=y\\ \end{array} $

Therefore, if $x \ne y$, $x-k$ and $y-k$ have different signs, so that either $x-k < 0$ and $y-k > 0$ or $x-k > 0$ and $y-k < 0$.

If $x=y$, then $0 = (x-k)(y-k) = (x-k)^2 $ so $x = y = k$.