Complicated Equation!

Using Cauchy-Schwarz inequality on the LHS:

$\begin{aligned} \left(\sqrt{1-2015x} + \sqrt{1+2015x}\right)^2 & \le (1+1)(1-2015x+ 1+2015x) = 4 \\ \sqrt{1-2015x} + \sqrt{1+2015x} & \le 2 \quad \quad \color{#3D99F6}{\text{Equality happens when }x=0} \end{aligned}$

Using AM-GM inequality on the RHS (note that RHS $\ge 0$ ):

$\begin{aligned} \frac{1}{\sqrt{x+1}} + \sqrt{1+x} & \ge 2 \sqrt{\frac{\sqrt{1+x}}{\sqrt{x+1}}} = 2 \quad \quad \color{#3D99F6}{\text{Equality happens when }x=0} \end{aligned}$

We note that the RHS $\le 2$ and LHS $\ge 2$ , therefore, there is only a point at $x=\boxed{0}$ , where RHS $=$ LHS $= 2$

Well my solution...

Why not use functions to solve this one. Let f be a function as $f(x)= \sqrt{1+x}$

Then the equation becomes $f(-2015x) + f(2015x) = \dfrac1{f(x)} + f(x)$

we know that $D_f = [-1 , + \infty [$ But in this case for the equation to be true $D_f = [ - \frac {1}{2015} ,\frac {1}{2015}]$ And the only possible solution is x = 0 So the sum of the soultions is 0

root(1/1+x) + root(1+x) >= 2 ;

(It implies) root(1+2015x) + root(1-2015x) >= 2 ;

Squaring both sides

1+2015x+1-2015x+2root(1-2015x^2)>=4 ;

root(1-2015x^2)>=1 ;

squaring both sides

Satisfying condition.

X=0