Why if $x>2 \sqrt x$, then $\sqrt x > 2$? I don't understand this step:
$$x>2 \sqrt x \Longrightarrow \sqrt x > 2$$
($x\geq 0 $)
 A: Divide by $\sqrt x$. Note that $x > 0$ holds as $\sqrt x$ is defined and for $x = 0$ we do not have $0 > 2 \sqrt 0$.
A: first I assume that $x> 0$
$$x>2 \sqrt x \Longrightarrow \sqrt x*\sqrt x > 2\sqrt x$$
now because $\sqrt x> 0$ you can divide both sides by $\sqrt x$ and you'll get
$$\sqrt x> 2$$
A: Hint: If $x > 0$, then $x = \sqrt{x}\cdot \sqrt{x}$
A: Yet another way. If $x>2\sqrt{x}$ we can square both sides since $x$ is non-negative. Then, $x^2>4x$ and thus $x(x-4)>0$. Since $x$ is non-negative, we must have $x>0$ and $x>4$. By putting roots on the $x>4$ we get $\sqrt{x}>2$.  This is not as fast as the other methodologies above, but I just put it to note another approach.
A: You are given
$$ x > 2 \sqrt{x}.$$
You can assume that $\sqrt{x}$ is defined. Otherwise the inequality doesn't make sense
Now, $\sqrt{x} \ge 0$ whenever $\sqrt{x}$ exists. But if $\sqrt{x} = 0$, then $x = 0$ and $2 \sqrt{x} = 0$, so $x \ngtr 2 \sqrt{x}$. Therefore $\sqrt{x} > 0$, strictly.
Since $\sqrt{x} > 0$, also $\frac 1{\sqrt{x}} > 0$. Therefore
$$ x \cdot \frac 1{\sqrt{x}}  > 2 \sqrt{x} \cdot \frac 1{\sqrt{x}}.$$
Simplifying,
$$ \sqrt{x} > 2.$$
