How to prove that $m+(4/m^2)>=0$ for every m greater or equal to 0 I have a problem with proving this simple theorem. I've already figured out that the best strategy is to factorise is, so that to get eg. a square or another expression that from the definition has to be non-negative, but have no idea how exactly should I do it. Could you help me guys?
 A: Let $m > 0$. Then we have that:
$$m^3 >0$$
which implies that:
$$m^3 + 4 > 0 $$
dividing both sides by $m^2$ you get:
$$m + \frac{4}{m^2} > 0$$
and you are done (for the case where $m=0$ you are dividing by 0).
A: Consider the following, if I have a positive real number and I add a positive real number to it, the result will be a positive real number (Closure under addition). Furthermore, if I take a positive real number and divide by a positive real number, the result will be a positive real number (Closure under multiplication). Also, if I square a real number it will be a positive real number (Closure under multiplication). Given those statements, you have m a positive real number plus 4/m^2 which is a positive number because it is the quotient of two positive numbers. Thus the result must be positive.
A: Apparently you needed to prove it for $m + (4/m^2) \ge 3$, given $m > 0$.
Multiply by $m^2$.
Take derivative: $3m^2 - 6m$, and check that there is a local top in $m = 0$ and local minimum at $m = 2$.
Given the positive sign of the $m^3$ term, you can plot the graph.
Fill in $m = 2$, to check that equality occurs, but it is a local minimum on $R^+$
A: By applying AM-GM inequality on $$\frac{m}{2},\frac{m}{2},\frac{4}{m^2}$$
We get 
$$\frac{\frac{m}{2}+\frac{m}{2}+\frac{4}{m^2}}{3}\ge \sqrt[3]{\frac{m}{2}\cdot\frac{m}{2}\cdot\frac{4}{m^2}}$$
$$\frac{m}{2}+\frac{m}{2}+\frac{4}{m^2} \ge 3$$
$$\Rightarrow m+\frac{4}{m^2} \gt 0$$
