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?

We can rewrite LHS as:
$$3[\frac 1b(\frac{a^2}{b^2})+\frac 1c(\frac{b^2}{c^2})+\frac 1a(\frac{c^2}{a^2})]\geq(\frac 1c+\frac 1b+\frac 1a)(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2})$$
This holds , due to Tchebychev inequality, if $a<b<c$, such that both factors on RHS are in ascending order of magnitude. Also we apply this inequality:
$$a^2c+b^2a+c^2b\geq3abc$$
Dividing both sides by $abc$ we get:
$$\frac ab+\frac bc+\frac ca\geq 3$$
Therefore:
$$\frac {a^2}{b^2}+\frac {b^2}{c^2}+\frac {c^2}{a^2}\geq 3$$
Therefore:
$$\frac 1b(\frac{a^2}{b^2})+\frac 1c(\frac{b^2}{c^2})+\frac 1a(\frac{c^2}{a^2})\geq(\frac 1c+\frac 1b+\frac 1a)$$
