Solve the inequality $\sqrt x+\sqrt{x+1}>\sqrt 3$. Solve the inequality $\sqrt x+\sqrt{x+1}&gt;\sqrt 3$.

I want to make sure my method is correct: 

The condition is that $x\geq 0$ 

$x+2\sqrt x\times \sqrt{x+1} +x+1&gt;3$

$2\sqrt{x(x+1)}&gt;2-2x$

$4x(x+1)&gt;4-8x+4x^{2}$

$4x^{2}+4x&gt;4-8x+4x^{2}$ 

$12x&gt;4$

$x&gt;\frac{1}{3}$ 

$x\in (\frac{1}{3},\infty)$

I know my final solution is fine, but is everything written properly? Should I put $\iff$ at the beginning of each row?
Over its domain (the set of non-negative real numbers) the function $f(x)=\sqrt{x}+\sqrt{x+1}$ is an increasing function, since it is the sum of two non-negative increasing functions. It follows that $f(x)&gt;\sqrt{3}$ holds as soon as $x&gt;x_0$, where $x_0$ is the only positive number such that
$$ \sqrt{x_0}+\sqrt{x_0+1} = \sqrt{3}.\tag{1} $$
$x_0=\frac{1}{3}$ is clearly a solution of $(1)$, hence the given inequality holds for $\color{red}{\large x&gt;\frac{1}{3}}$.
Note that $\sqrt{\frac13}+\sqrt{\frac13+1}=\sqrt{3}$ . The inequality  follows trivially.