# Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$

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.

Thanks :)

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

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

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

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 $a=b=c.$ $\Box$

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.

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.

Let $$M=\dfrac{a^3}{a+b}+\dfrac{b^3}{b+c}+\dfrac{c^3}{c+a}$$ be the original expression, we introduce its "conjugate" $$N=\dfrac{ab^2}{a+b}+\dfrac{bc^2}{b+c}+\dfrac{ca^2}{c+a}$$ Direct computing, and using $a^2+b^2+c^2\geq ab+bc+ca$, yields $$M-N =\frac{a(a^2-b^2)}{a+b}+\frac{b(b^2-c^2)}{b+c}+\frac{c(c^2-a^2)}{c+a}\\=a(a-b)+b(b-c)+c(c-a)\\= a^2+b^2+c^2-ab-bc-ca\geq 0$$ Therefore $M\geq N$. Also using $2(x^2+y^2)\geq (x+y)^2$ $$M+N=\frac{a(a^2+b^2)}{a+b}+\frac{b(b^2+c^2)}{b+c}+\frac{c(c^2+a^2)}{c+a}\\ \geq \frac{a(a+b)+b(b+c)+c(c+a)}{2} \\ =\frac{a^2+b^2+c^2+ab+bc+ca}{2} \\ \geq ab+bc+ca$$ Hence we have both $M\geq N$ and $M+N\geq ab+bc+ca$. It follows that $M\geq \dfrac{ab+bc+ca}{2}$.

By Holder $$\sum_{cyc}\frac{a^3}{a+b}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}(a+b)}=\frac{(a+b+c)^2}{6}\geq\frac{ab+ac+bc}{2}.$$ Done!

The Holder for three sequences it's the following.

Let $a_i$,$b_i$, $c_i$, $\alpha$, $\beta$ and $\gamma$ be positive numbers. Prove that: $$(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_n)^{\beta}(c_1+c_2+...+c_n)^{\gamma}\geq$$$$\geq\left(\left(a_1^{\alpha}b_1^{\beta}c_1^{\gamma}\right)^{\frac{1}{\alpha+\beta+\gamma}}+\left(a_2^{\alpha}b_2^{\beta}c_2^{\gamma}\right)^{\frac{1}{\alpha+\beta+\gamma}}+...+\left(a_n^{\alpha}b_n^{\beta}c_n^{\gamma}\right)^{\frac{1}{\alpha+\beta+\gamma}}\right)^{\alpha+\beta+\gamma}.$$

In our case $\alpha=\beta=\gamma=1$, $a_1=\frac{a^3}{a+b},$ $a_2=\frac{b^3}{b+c}$, $a_3=\frac{c^3}{c+a},$

$b_1=a+b,$ $b_2=b+c$, $b_3=c+a$ and $c_1=c_2=c_3=1$.

• No idea how the holder's inequality applies here. Can you give a detailed proof? – Mathis Dec 14 '17 at 11:56
• @Mathis I added something. See now. – Michael Rozenberg Dec 14 '17 at 12:09