Let $a;c>0$ such that $a^2+2ac=1$. Prove that $$2\sqrt{\frac{a}{a+c}}-\frac{9\sqrt{c^2+1}}{8c}<1$$
My proof:
$\Leftrightarrow 2\sqrt{\frac{a}{a+c}}-\frac{9\sqrt{c^2+a^2+2ac}}{8c}<1$
$\Leftrightarrow 2\sqrt{\frac{a}{a+c}}<1+\frac{9\left(a+c\right)}{8c}$
$\Leftrightarrow \frac{4a}{a+c}<\frac{81a^2+306ac+289c^2}{64c^2}$
$\Leftrightarrow \frac{81a^3+387a^2c+339ac^2+289c^3}{64c^2\left(a+c\right)}>0$
The last inequality is true by $a;c>0$. I want some other way and a inequality is rigider than original inequality. Tks
$2\sqrt{\frac{a}{a+c}}-\frac{9\sqrt{c^2+1}}{8c}=2\sqrt{\frac{a+c-c}{a+c}}-\frac{9\sqrt{a^2+c^2+2ac}}{8c}=2\sqrt{1-\frac c{a+c}}-\frac{9(a+c)}{8c}<2-\frac{9c}{8c}=2-\frac98=\frac78<1$