1. ## Equation

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

2. Hello,

If you're trying to prove that there is a unique solution, then your method is good. And it's easy to prove because $\displaystyle \displaystyle f(x)=\frac{1}{\sqrt{1-x}}+\frac{4}{1+2x}$

If you want to find a, then that's more complicated I guess can you first confirm what you are looking for ?

3. I want to find a; if f is injective/ strictly monotone, then $\displaystyle 1-a=a \Leftrightarrow a=\frac{1}{2}$ which verify the equation.

4. Well, let u = 3a - 2a^2 and v = a + 1 - 2a^2, then:

v[3 + 14a +8sqrt(u)] = u[17 - 14a +8sqrt(v)]

5. $\displaystyle (\sqrt{v}-\sqrt{u})[8\sqrt{uv}-7(u+v)(\sqrt{u}+\sqrt{v})+10(\sqrt{u}+\sqrt{v})]=0 \Rightarrow$
I. $\displaystyle \sqrt{v}-\sqrt{u}=0 \Rightarrow a=\frac{1}{2}$
II. $\displaystyle 8\sqrt{uv}-7(u+v)(\sqrt{u}+\sqrt{v})+10(\sqrt{u}+\sqrt{v})=0 \Rightarrow 8\sqrt{uv}+(\sqrt{u}+\sqrt{v})[7(2a-1)^{2}-4]=0$
And now?
$\displaystyle 8\sqrt{uv}\geq 0$, $\displaystyle (\sqrt{u}+\sqrt{v})\geq 0$, but $\displaystyle 7(2a-1)^{2}-4 \epsilon (-4, 3)$