$\displaystyle \frac{\sqrt{3-2a}+4\sqrt{a}}{\sqrt{1+2a}+4\sqrt{1-a}}=\frac{\sqrt{-2a^{2}+3a}}{\sqrt{-2a^{2}+a+1}}$, a is a real number.

What I did:

$\displaystyle \frac{\sqrt{3-2a}+4\sqrt{a}}{\sqrt{1+2a}+4\sqrt{1-a}}=\frac{\sqrt{-2a^{2}+3a}}{\sqrt{-2a^{2}+a+1}} \Leftrightarrow \frac{\sqrt{3-2a}+4\sqrt{a}}{\sqrt{-2a^{2}+3a}}=\frac{\sqrt{1+2a}+4\sqrt{1-a}}{\sqrt{-2a^{2}+a+1}}$

$\displaystyle \Leftrightarrow \frac{\sqrt{1+2(1-a))}+4\sqrt{1-(1-a)}}{\sqrt{[1+2(1-a)][1-(1-a)]}}=\frac{\sqrt{1+2a}+4\sqrt{1-a}}{\sqrt{(1+2a)(1-a)}}

$

Let $\displaystyle \displaystyle f0, 1)\mapsto \mathbb{R}, f(x)=\frac{\sqrt{1+2x}+4\sqrt{1-x}}{\sqrt{(1+2x)(1-x)}}\Rightarrow \displaystyle f(1-a)=f(a)$. I want to prove that f is injective/ strictly monotone.

Well, that is my idea, but anything else would be okay. Thanks in advance.