Prove $\sqrt{5} \leq 3$

Squaring both sides of an inequality does not preserve the inequality unless both sides were originally positive.

The long way: suppose $\sqrt{5} > 3$. Multiplying both sides by $3$, we have $3\sqrt{5} > 9$. Since $3 < \sqrt{5}$, $3\sqrt{5} < \sqrt{5}^2 = 5$. So $5 > 3\sqrt{5} > 9$, a contradiction.

Here is another way to look at this problem.

The square-root function, $f(x)=\sqrt x$, is increasing on its domain because $f'(x)>0$ on $(0, +\infty)$. Since $3=\sqrt 9$,

$3=f(9)$.

And $\sqrt 5=f(5)$.

But since $f(x)=\sqrt x$ is increasing on its domain, by the definition of increasing (and because $5<9$):

$f(5) <f(9)$.

Substitute the meanings of $f(5)$ and $f(9)$ and get:

$\sqrt 5 <3$.

Q.E.D.

Hint: $\;\;3-\sqrt{5} \,=\, \cfrac{4}{3+\sqrt{5}}$