How to show that $\inf({\frac{mn}{1+m+2n}}) = 1/4$? I am trying to prove the result:
$\inf({\frac{mn}{1+m+2n}}) = \frac{1}{4}$, where $m, n \in\mathbb{N}$.
I proved existence of infimum, but then I do not know how to prove that $1/4$ is a lower bound and then can not prove that $1/4$ is the greatest lower bound.

To show it achieves this lower bound, let $m = n = 1$. Then
$$\frac{mn}{1+m+2n} = {1 \over 1+1+2} = \frac{1}{4}.$$
Now we need to show for all other $m,n \in \Bbb N$,
$$\frac{mn}{1+m+2n} \ge \frac14.$$
If $n \ge 2$, then
$$\frac{mn}{1+m+2n} \ge \frac{2m}{5+m} \ge \frac{2}{5+1} = \frac{1}{3} > \frac14.$$
If $m \ge 2$, then
$$\frac{mn}{1+m+2n} \ge \frac{2n}{3+2n} \ge \frac{2}{3+2} = \frac25 > \frac14.$$
Thus we have shown it holds for all cases.

Note: We used the fact that $\frac{2x}{5+x}$ and $\frac{2x}{3+2x}$ are increasing functions for $x \ge 0$.
To prove this, let $0 \le x < y$. Then
$$10 x < 10 y$$
$$10 x + 2xy < 10y + 2xy$$
$$2x(5+y) < 2y(5+x)$$
$$\frac{2x}{5+x} < \frac{2y}{5+y}$$
which shows $\frac{2x}{5+x}$ is increasing. The proof for $\frac{2x}{3+2x}$ is analogous.
