If 1 $\leq x$, then $\sqrt{x} \leq x $ This is a really simple problem but I am unsure if I have proved it properly.
By contradiction:
Suppose that $x \geq 1$ and $x< \sqrt{x}$. Then $x\cdot x \geq x \cdot 1$ and $x^2 < x$ (squaring both sides), which is a contradiction.
 A: Here is a direct proof (without contradiction)
$x=1+r$ with $r\geq0$ . Then $x^2=(1+r)^2=1+2r+r^2\geq1+r=x\rightarrow x^2\geq x\rightarrow x\geq\sqrt x$
A: Assume $x \geq 1.$ Then $x - 1 \geq 0$ and $x > 0,$ and hence
$$ x(x-1) \; = \; x^2 - x \; \geq \; 0$$
since the product of two non-nonegative expressions is non-negative.
Factoring $x^2 - x$ as a difference of squares gives
$$ \left(x - \sqrt{x} \right) \left( x + \sqrt{x} \right) \; \geq \; 0$$
Since $x + \sqrt{x}$ is positive (both $x$ and $\sqrt{x}$ are positive), it follows that $x - \sqrt{x}$ is non-negative, which is easily seen to be equivalent to what you wanted to prove.
A: As a contra-positive, assuming $x$ is not negative:
$$\sqrt{x} > x \implies x>x^2 \land x \neq0 \implies 1>x$$
(1st implication is by squaring (Which is obviously an increasing function here), 2nd implication is by dividing by $x$)  
So, equivalently:
$$x \geq 1  \implies x \geq\sqrt{x}$$
Alternatively, for a more direct proof, you can rely on $\sqrt{\cdot}$ being an increasing function in $\mathbb R_{\ge 0}$ domain: 
$$x \geq 1 \implies  x^2 \geq x \implies x \geq\sqrt{x}$$
(1st implication is by multiplying by $x$, 2nd implication is by taking the square root, relying on it being increasing)
