Calculation of real root values of $x$ in $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}.$ 
Calculation of x real root values from $ y(x)=\sqrt{x+1}-\sqrt{x-1}-\sqrt{4x-1} $

$\bf{My\; Solution::}$ Here domain of equation is  $\displaystyle x\geq 1$. So squaring both sides we get
$\displaystyle (x+1)+(x-1)-2\sqrt{x^2-1}=(4x-1)$.
$\displaystyle (1-2x)^2=4(x^2-1)\Rightarrow 1+4x^2-4x=4x^2-4\Rightarrow x=\frac{5}{4}.$
But when we put $\displaystyle x = \frac{5}{4}\;,$ We get $\displaystyle \frac{3}{2}-\frac{1}{2}=2\Rightarrow 1=2.$(False.)
So we get no solution.
My Question is : Can we solve above question by using comparision of expressions?
Something like $\sqrt{x+1}<\sqrt{x-1}+\sqrt{4x-1}\; \forall x\geq 1?$ 
If that way possible, please help me solve it. Thanks.

Also, for $x\ge1$ $$\sqrt{4x-1}+\sqrt{x-1}\ge\sqrt{4x-1+x-1}=\sqrt{5x-2}>\sqrt{2x}\ge\sqrt{x+1}. $$
