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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?

Barry Cipra
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user
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4 Answers4

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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.

user133281
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  • Can you tell me how you found the required coefficients? I always see these sorts of solutions and can't help but think that they come out of thin air. – Oussema Mar 06 '21 at 16:57
  • @Oussema you can reverse engineer them and solve a linear system of equation with three variables. – dezdichado Mar 06 '21 at 19:25
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There are million ways to prove this but the most immediate one would be the Rearrangement Inequality. No matter the order of $a,b,c,$ the triplets $\left(\dfrac{1}{a^3}, \dfrac{1}{b^3}, \dfrac{1}{c^3}\right)$ and $\left(a^2, b^2, c^2\right)$ have the reverse arrangement.

dezdichado
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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:

$$\frac 1b(\frac{a^2}{b^2})+\frac 1c(\frac{b^2}{c^2})+\frac 1a(\frac{c^2}{a^2})\geq(\frac 1c+\frac 1b+\frac 1a)$$

sirous
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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.

Jan Safronov
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