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?
 A: For quite a different approach, you could differentiate $f(m)=m+\frac{4}{m^2}$ w.r.t. $m$, resulting in $f'(m)=1-\frac{8}{m^3}$. 
Setting $f'(m)$ to zero and solving for $m$ results in $m=2$, so that the minimum is $3$ (given that the second derivative $f''(m)=\frac{24}{m^4}$ is positive for $m=2$). 
A: Or if you want different approach from AM-GM inequality 
$m^3-3m^2+4=m^3-2m^2-m^2+4=m^2(m-2)-(m-2)(m+2)=(m-2)(m^2-m-2)=(m-2)^2\cdot (m+1)$ 
since $m+1>0$ , and $(m-2)^2 \geq 0$ , we are done. Equality occurs when $m=2$
A: Hint: use $a+b+c\ge 3\sqrt[3]{abc}$ for $a=b=\frac m{2},c=\frac{4}{m^2}$.
A: 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
$$m + \frac{4}{m^2}\geq 3 ~ \forall m > 0$$
A: $$m^3-3m^2+4\geq 0$$
now we assume the $$f(m)=m^3-3m^2+4$$
the  minimum value of function $f(m)$ as follows
$$f'(m)=3m^2-6m=0$$
$$3m(m-2)=0$$ 
$$m_1=0$$
$$m_2=2$$
the $m_2=2$ gives the minimum value which is $f(2)=0$ if $m\geq0$, 
so the $f(m)$ must be equal or greater than $0$
A: 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 $m=2$
