Prove that $K(x)< \frac{\pi}{2}\frac{1}{\sqrt{1-0.53 x}}$, $\forall x\in(0,0.25)$

Question

Does anyone have an idea on how to prove the following inequality? $$K(x)< \frac{\pi}{2}\frac{1}{\sqrt{1-0.53 x}},\quad\forall x\in(0,0.25)$$ where $$K(x)=\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{1-x \sin^2\theta}} d\theta$$

Answer 1

It's known (cf. Wikipedia) that $$K(x)=\frac{\pi}{2\operatorname{agm}(1,\sqrt{1-x})},$$ where the arithmetic-geometric mean $\operatorname{agm}(a,b)$ is defined as the common limit of two sequences defined via $$a_0=a, \quad b_0=b$$ and $$a_{n+1}=\frac{a_n+b_n}2,\quad b_{n+1}=\sqrt{a_nb_n}$$ for $n\ge0.$ It's clear that $b_n\le \operatorname{agm}(a,b)\le a_n$ for all $n$. In our case, we have$$b_2=\sqrt{\frac{(1-x)^{1/4}+(1-x)^{3/4}}2}.$$ Now the function $$f(x)=\frac{(1-x)^{1/4}+(1-x)^{3/4}}2$$ is concave on $[0,1]$, so for $x\in[0,x_0],$its graph lies above the line joining $(0,f(0))$ and $(x_0, f(x_0))$: $$f(x)\ge 1-cx,$$ where $c=(1-f(x_0))/x_0.$ For $x_0=0.25,$ we get $c=0.526935384\ldots<0.53.$