Cyclic Inequality via AM-GM and Cauchy-Schwarz

This problem is a beautiful example of a cyclic inequality involving rational expressions. The challenge is to bridge the gap between awkward-looking fractions and the clean symmetric form on the right. Two distinct methods are presented: one using AM-GM with a clever pairing trick to cancel denominators, and an alternative using Cauchy-Schwarz that rewrites the whole problem in a single elegant step.

Problem Statement

If $a, b, c > 0$, prove that:

$$\frac{a^3}{b} + \frac{b^3}{c} + \frac{c^3}{a} \ge ab + bc + ca$$

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Hints

• Hint 1: You need to eliminate the denominators. What existing term from the right-hand side can you pair with $\frac{a^3}{b}$ so that applying the AM-GM inequality simplifies nicely to an $a^2$ term?
• Hint 2: Keep the standard inequality $a^2 + b^2 + c^2 \ge ab + bc + ca$ in your back pocket; you will need it to bridge the final gap.
• Stuck on AM-GM? There is an elegant alternative solution using the Cauchy-Schwarz inequality!

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Solutions

Method 1: Using AM-GM

Apply the AM-GM inequality to cleverly chosen pairs to cancel the denominators:

$$\frac{a^3}{b} + ab \ge 2\sqrt{\frac{a^3}{b} \cdot ab} = 2a^2$$

Apply this cyclically for the $b$ and $c$ terms, then sum the three inequalities:

$$\left( \frac{a^3}{b} + \frac{b^3}{c} + \frac{c^3}{a} \right) + (ab + bc + ca) \ge 2(a^2 + b^2 + c^2)$$

Since we know $a^2 + b^2 + c^2 \ge ab + bc + ca$, we can substitute the right side:

$$\left( \frac{a^3}{b} + \frac{b^3}{c} + \frac{c^3}{a} \right) + (ab + bc + ca) \ge 2(ab + bc + ca)$$

Subtract $(ab + bc + ca)$ from both sides to complete the proof. $\square$

Method 2: Using Cauchy-Schwarz (Alternative)

Multiply the left-hand side by $(ab + bc + ca)$ and apply Cauchy-Schwarz:

$$\left( \frac{a^3}{b} + \frac{b^3}{c} + \frac{c^3}{a} \right)(ab + bc + ca) \ge \left( \sqrt{a^4} + \sqrt{b^4} + \sqrt{c^4} \right)^2 = (a^2 + b^2 + c^2)^2$$

Using the known inequality $a^2 + b^2 + c^2 \ge ab + bc + ca$, we can square both sides to get $(a^2 + b^2 + c^2)^2 \ge (ab + bc + ca)^2$. Substituting this into the right side of our Cauchy-Schwarz result:

$$\left( \frac{a^3}{b} + \frac{b^3}{c} + \frac{c^3}{a} \right)(ab + bc + ca) \ge (ab + bc + ca)^2$$

Divide both sides by the strictly positive term $(ab + bc + ca)$ to complete the proof. $\square$

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Takeaways

• Strategic Pairing: In AM-GM, adding a term specifically to cancel a denominator is a common and powerful trick for cyclical inequalities.
• Stepping-Stone Inequalities: Famous identities like $a^2 + b^2 + c^2 \ge ab + bc + ca$ frequently act as a bridge between your algebraic manipulation and the final required format.
• Multiple Paths: Inequality problems often have multiple viable vectors of attack. If AM-GM feels clunky, reshaping the equation to fit Cauchy-Schwarz can sometimes provide a much faster route.

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Further Readings

HSC Induction, HSC Sequences