Prove that $m+\frac{4}{m^2}\ge3$ How to prove that $m+\frac{4}{m^2}\geq3$ for every $m>0$? I multiplied both sides by $m^2$ and finally got $m^3+4-3m^2\geq0$, yet don't know how to move further. Hint?

If $m=1$, then
$$m + \frac{4}{m^2}= 1 + \frac{4}{1^2} = 1 + 4 = 5 \geq 3$$
If $m = 2$, then 
$$m + \frac{4}{m^2} = 2 + \frac{4}{2^2} = 2 + 1 = 3 \geq 3$$
Consider now $m\geq 3$. Then, since $\frac{4}{m^2}$ is positive for all $m$, it is obvious that also $m + \frac{4}{m^2}$ is greater than $3$.
We can conclude that
$$m + \frac{4}{m^2}\geq 3 ~ \forall m > 0$$
