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How to deal with $m+\frac{4}{m^2}\geq3$ for every $m > 0$ ? I multiplied both sides by $m^2$ and got $m^3+4-3m^2\geq0$ and have no idea how to continue with this

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$$\frac{m}{2} + \frac{m}{2}+ \frac{4}{m^2} \geq 3(\frac{m \cdot m \cdot 4}{2\cdot 2 \cdot m^2})^{1/3} = 3$$

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we we have the inequality $m^3+4-3m^2>0$ we see that $m=-1$ and $m=2$ are solutions of $m^3-3m^2+4$ thus we can write $(m+1)(m-2)^2$ and this is not negative. $m=2$ is a double solution of $m^3-3m^2+4=0$

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