<s>[INST] 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?
 [/INST] Problems of this type can often be solved by showing, using AM-GM, that each term on the right hand side is not greater than some linear combination of the terms on the left hand side.
In this particular situation,
$$
4\cdot \frac{a^2}{b^3} + 6\cdot \frac{b^2}{c^3} + 9\cdot\frac{c^2}{a^3} \ge 19 \cdot \sqrt[19]{\left(\frac{a^2}{b^3}\right)^4 \left(\frac{b^2}{c^3}\right)^6 \left(\frac{c^2}{a^3}\right)^9} = 19 \cdot \sqrt[19]{\frac{a^8}{a^{27}}} = 19 \cdot \frac{1}{a}.
$$
Adding this inequality to its cyclic variants yields the result.
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