# How to prove that $m+(4/m^2)>=0$ for every m greater or equal to 0 [duplicate]

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?

## marked as duplicate by Jyrki Lahtonen, user147263, Najib Idrissi, user99914, HakimNov 2 '14 at 10:52

• Maybe prove $m^3 + 4 \geq 0$ instead? – IAmNoOne Nov 1 '14 at 22:30
• You know that $m\ge0$, and $m^2>0$ (assuming $m\ne0$), so $\frac{4}{m^2}>0$ also. Thus $m+\frac{4}{m^2}>0$ for $m>0$. – user84413 Nov 1 '14 at 22:33
• For $m=0$: $\dfrac{4}{m^2}$ doesn't exist. – Vladimir Vargas Nov 1 '14 at 22:33
• Thanks! Actually, how to deal with $m+(4/m^2)>=3$ for every m >= 0 ? Of course I multiplied it by $m^2$ and got $m^3+4-3m^2>=0$ and have no idea how to proceed – applicant Nov 1 '14 at 22:38
• If the right side of your inequality should be $3$ you should immediately edit your question's title to ask the correct question. Otherwise you are wasting the time of people who are trying to help you. Also note Vladimir's comment that your statement cannot be true for $m=0$, so you should edit your restriction on $m$. – Rory Daulton Nov 1 '14 at 22:58

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).

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.

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 $R^+$

By applying AM-GM inequality on $$\frac{m}{2},\frac{m}{2},\frac{4}{m^2}$$ We get $$\frac{\frac{m}{2}+\frac{m}{2}+\frac{4}{m^2}}{3}\ge \sqrt[3]{\frac{m}{2}\cdot\frac{m}{2}\cdot\frac{4}{m^2}}$$

$$\frac{m}{2}+\frac{m}{2}+\frac{4}{m^2} \ge 3$$ $$\Rightarrow m+\frac{4}{m^2} \gt 0$$