Prove the arithmetic geometric mean inequality. That is, for two positive real numbers x,y we have sqrt(xy) is less than or equal to (x+y)/2. Furthermore, equality occurs if and only if x = y.
I have proved the first part but I was wondering if someone could show me how to prove the second part. I can show how I proved the first part... if that is needed?
Hint: If $\sqrt{xy} = \frac{x + y}{2}$, square both sides and do some algebra to conclude that $(x - y)^2 = 0$.
So you need to show $\sqrt{xy}=\frac{x+y}{2}\iff x=y$
proof of $\implies$:
$\sqrt{xy}=\frac{x+y}{2}$. Squaring both sides gives $xy=\frac{(x+y)^2}{4}.$ After simplification, we have $x^2-2xy+y^2=0$ which factors into $(x-y)^2=0$. From this we see that $x=y$.
proof of $\impliedby:$ Since $\sqrt{x^2}=x$, and $x$ can be written as $x=\frac{x+x}{2}$, then we have $\sqrt{xx}=\frac{x+x}{2}$. Since by assumption $x=y$, then we have that $\sqrt{xy}=\frac{x+y}{2}.\quad\quad\square$
