Prove $\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}≥\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ Question : Prove $$\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
$(a, b, c \in \mathbb{R}^+)$
I tried to solve it like this :
$$\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \; (\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2})$$
Am I doing this right? How can I finish this problem?

It can be proven in a straightforward way, just basic inequalities and definition of the real numbers, we'll have the inequality: $0\leq a\leq b\leq c$.
$$c^2\geq a^2 \Rightarrow \frac{c^2}{a^3}\geq\frac{1}{a}$$
$$\text{and} \quad (\frac{c}{b})^n \geq0> \frac{c^2-b^2}{a^2-b^2} \Rightarrow c^3(a^2-b^2)\geq b^3(c^2-b^2)\Rightarrow c^3a^2-c^3b^2\geq b^3c^2-b^5 \\ a^2-b^2\geq \frac{b^3}{c} - \frac{b^5}{c^3} \Rightarrow \frac{a^2}{b^3}-\frac{1}{b}\geq\frac{1}{c}-\frac{b^2}{c^3} \\ \frac{a^2}{b^3}+\frac{b^2}{c^3}\geq\frac{1}{b}+\frac{1}{c}$$ 
You can see the same thing happening for any order of inequalities between $a,b,c$, and you combine the two equations from above.
