# Bounding a ratio of arithmetic geometric means

Let $\alpha$ be a real number in $[1/2,5/6]$.

How do I easily prove that $$AGM(1,\sqrt{1-\alpha})\leq AGM(1,\sqrt{\alpha}) \leq 3AGM(1,\sqrt{1-\alpha})?$$

AGM denotes the arithmetic geometric mean.

-
Arithmetic geometric mean $M(1,x)$ for $x>0$ is related to complete elliptic integral: $$M(1,x) = \frac{\pi (x+1)}{4 K\left(\frac{(1-x)^2}{(1+x)^2}\right)}$$ From the definition of $M(1,x)$ is clear that for $0<x<1$, $0<M(1,x)<1$ and that $M(1,x)$ is increasing.
Thus to verify your inequality it suffices to check end-points. The left inequality is saturated for $\alpha=\frac{1}{2}$, and so the right inequality follows since $M(1,x)>0$ for $x = 2^{-1/2}$. Let $m_1 = M\left(1,\frac{1}{\sqrt{6}}\right)$ and $m_2 = M\left(1,\sqrt{\frac{5}{6}}\right)$, but $$m_1 = 0.6711 \qquad m_2 = 0.9559 \qquad \implies \qquad m_1 < m_2 < 3 m_1$$ From this analysis it follows that the inequality can be strengthened to : $$\operatorname{AGM}(1,\sqrt{1-\alpha})\leq \operatorname{AGM}(1,\sqrt{\alpha}) \leq \frac{3}{2} \operatorname{AGM}(1,\sqrt{1-\alpha})$$