Why if $x>2 \sqrt x$, then $\sqrt x > 2$?

Question

I don't understand this step:

$$x>2 \sqrt x \Longrightarrow \sqrt x > 2$$

($x\geq 0 $)

Answer 1

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

$$\sqrt x> 2$$

Answer 2

Divide by $\sqrt x$. Note that $x > 0$ holds as $\sqrt x$ is defined and for $x = 0$ we do not have $0 > 2 \sqrt 0$.

Answer 3

Hint: If $x > 0$, then $x = \sqrt{x}\cdot \sqrt{x}$

Answer 4

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.

Answer 5

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, $$ \sqrt{x} > 2.$$