<s>[INST] 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?
 [/INST] Using $A.M\geq G.M$
$$\frac{\frac{m}{4}+\frac{m}{4}+\frac{m}{4}+\frac{m}{4}+\frac{m}{4}+\frac{2}{m^2}+\frac{2}{m^2}}{6}\geq\frac{1}{2}$$
$$m+\frac{4}{m^2}\geq3$$
Equality occurs if  
$$\frac{m}{4}=\frac{m}{4}=\frac{m}{4}=\frac{m}{4}=\frac{m}{4}=\frac{2}{m^2}=\frac{2}{m^2}$$
$i.e.$ for $m=2$
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