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.
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.