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prove the following inequality:

$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2},$$ for $a,b,c$ real positive numbers.

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4 Answers 4

By the Cauchy-Schwarz Inequality: $$\frac{a^4}{a^2+ab} + \frac{b^4}{b^2 + bc} + \frac{c^4}{c^2 + ac} \ge \frac{(a^2+b^2+c^2)^2}{a^2 + b^2 + c^2 + ab + ac + bc}$$ Now, I claim that $$\frac{(a^2+b^2+c^2)^2}{a^2 + b^2 + c^2 + ab + ac + bc} \ge \frac{ab+ac+bc}{2}$$ Expanding, it suffices to show: $$2a^4 + 2b^4 + 2c^4 + 4a^2b^2 + 4a^2c^2 + 4b^2c^2 \ge a^3b + a^3c + b^3a + b^3c + c^3a + c^3b + 3a^2bc + 3b^2ac + 3c^2ab + a^2b^2 + a^2c^2 + b^2c^2$$ Using the inequality $\frac{3}{4}a^4 + \frac{1}{4}b^4 \ge a^3b$ which is a consequence of AM-GM, we can show: $$2a^4 + 2b^4 + 2c^4 \ge a^3b + a^3c + b^3a + b^3c + c^3a + c^3b$$ Thus it suffices to show: $$4a^2b^2 + 4a^2c^2 + 4b^2c^2 \ge 3a^2bc + 3b^2ac + 3c^2ab + a^2b^2 + a^2c^2 + b^2c^2$$ Which is simply just: $$3a^2b^2 + 3a^2c^2 + 3b^2c^2 \ge 3a^2bc + 3b^2ac + 3c^2ab$$ By AM-GM one has $\frac{1}{2}a^2b^2 + \frac{1}{2}a^2c^2 \ge a^2bc$. From applying this a bunch one easily gets the desired result.

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up vote 1 down vote accepted

$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} =\frac{a^4}{a^2+ab}+\frac{b^4}{b^2+bc}+\frac{c^4}{c^2+ac}\geq \frac{(a^2+b^2+c^2)^2}{a^2+b^2+c^2+bc+ca+ab}.\tag{1}$$ But $$a^2+b^2+c^2 \geq ab+bc+ca $$ so

$$a^2+b^2+c^2+ab+bc+ca \leq 2(a^2+b^2+c^2)$$

$$(1) \geq \frac{(a^2+b^2+c^2)^2}{2(a^2+b^2+c^2)} \geq \frac{ab+bc+ca}{2}.$$

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I started this proof but then realized that there was a mistake. Below is the wrong argument. Enjoy figuring out the mistake. (The argument can be fixed though.) \begin{align} S(a,b,c) & = \dfrac{a^3}{a+b} + \dfrac{b^3}{b+c} + \dfrac{c^3}{c+a}\\ & = \dfrac{a^3+b^3}{a+b} + \dfrac{b^3+c^3}{b+c} + \dfrac{c^3+a^3}{c+a} - \left(\dfrac{b^3}{a+b} + \dfrac{c^3}{b+c} + \dfrac{a^3}{c+a} \right)\\ & = (a^2 - ab + b^2) + (b^2 - bc + c^2) + (c^2 - ca + a^2) - S(a,c,b) \end{align} Hence, $$S(a,b,c) + S(a,c,b) = (a^2 - ab + b^2) + (b^2 - bc + c^2) + (c^2 - ca + a^2) \geq ab + bc+ ca$$ If $S(a,b,c) < \dfrac{ab + bc+ ca}2$, then $S(a,c,b) < \dfrac{ac + cb+ ba}2$ which gives us $$S(a,b,c) + S(a,c,b) < ab + bc + ca$$ Contradiction.

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Note that using the AM-GM inequality, we have $$\frac{a^3}{a+b}=a^2-\frac{a^2b}{a+b}\geq a^2-\frac{a^2b}{2\sqrt{ab}}=a^2-\frac 12\sqrt{a^3b},$$ And so summing up cyclically, it is enough to check that $$2(a^2+b^2+c^2)\geq\sqrt{a^3b}+\sqrt{b^3c}+\sqrt{c^3a}+ab+bc+ca,$$ Which is obviously true since $\sqrt{a^3b}\leq \dfrac{a^2+ab}2$ and $ab+bc+ca\leq a^2+b^2+c^2$ from the AM-GM inequality. Equality holds iff $a=b=c.$ $\Box$

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