November
2, 2006
If $x$
and $y$ are positive numbers, show that
$\displaystyle\sqrt{\frac{\left(\frac{x}1
\right)^2+\left(\frac{x+y}{2} \right)^2}{2}} \le
\frac{\sqrt{\left(\frac{x}1 \right)^2} + \sqrt{\frac{x^2+y^2}{2}}}{2}$
and find all cases of equality.
Solution. The inequality
may be rewritten as
$\sqrt{(2x)^2+(x+y)^2} \le
\sqrt{2}\,x+\sqrt{x^2+y^2}$.
We square both sides in order to rid ourselves of some radicals. This
gives us
$5x^2+2xy+y^2 \le 2x^2+
2\sqrt{2}\,x\sqrt{x^2+y^2}+x^2+y^2$.
Cancellation leads to
$x+y\le \sqrt{2(x^2+y^2)}$.
Squaring once again gives
$x^2+2xy+y^2\le 2(x^2+y^2) \quad\Longleftrightarrow
\quad 0\le (x-y)^2$.
All of these inequalities are equivalent. In the last, we easily see
that equality occurs only when $x=y$, so that the same is true of our
original inequality.