# prove $\sum_{cyc}\frac{a^3}{b}\ge ab+bc+ca$ if $a,b,c>0$

prove $$\sum_\text{cyc}\frac{a^3}{b}\ge ab+bc+ca$$ if $$a,b,c>0$$

I couldn't proceed much. I tried rearranging the inequality and it became

$$a^4c+b^4a+c^4b\ge a^2b^2c+b^2c^2a+c^2a^2b.$$

I tried using SOS here but it did not work.Also assuming $$a\ge b\ge c$$ didn't make things easier.

I also tried to work with one variable but it is a fourth degree so I skipped the calculus approach. We are actually supposed to prove using A-M G-M but other methods are also welcome.

Also, SOS helps here: $$\sum_{cyc}\left(\frac{a^3}{b}-ab\right)=\sum_{cyc }\frac{a(a^2-b^2)}{b}=$$ $$=\sum_{cyc}\left(\frac{a(a^2-b^2)}{b}-(a^2-b^2)\right)=\sum_{cyc}\frac{(a-b)^2(a+b)}{b}\geq0.$$

• i was able to follow all the solutions except the rearrangement one. I have not learned it yet.could you (if possible) explain a bit more on that one Thanks!! Commented Aug 26, 2020 at 4:34
• @Quantum Explain about Rearrangement? Commented Aug 26, 2020 at 4:36
• yes i dont know about that method Commented Aug 26, 2020 at 4:37
• never mind i researched it up a bit.I understand it now Thanks for the effort taken Commented Aug 26, 2020 at 4:58
• @Quantum You are welcome! Commented Aug 26, 2020 at 4:59

You can not assume that $$a\geq b\geq c$$ because the inequality is cyclic and not symmetric.

Since $$(a^3,b^3,c^3)$$ and $$\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$$ have an opposite ordering, by Rearrangement we obtain: $$\sum_{cyc}\frac{a^3}{b}\geq\sum_{cyc}\frac{a^3}{a}=\sum_{cyc}a^2\geq\sum_{cyc}ab,$$ where the last inequality it's also Rearrangement or it's just $$\frac{1}{2}\sum_{cyc}(a-b)^2\geq0.$$ About Rearrangement see here: https://math.stackexchange.com/edit-tag-wiki/5774

In our case triples $$(a^3,b^3,c^3)$$ and $$\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$$ have an opposite ordering, which says $$\sum_{cyc}\frac{a^3}{b}=\sum_{cyc}a^3\cdot\frac{1}{b}\geq \sum_{cyc}a^3\cdot\frac{1}{a}=\sum_{cyc}a^2.$$ Now, triples $$(a,b,c)$$ and $$(a,b,c)$$ have the same ordering, which says: $$\sum_{cyc}a^2=\sum_{cyc}a\cdot a\geq\sum_{cyc}ab.$$

$$a(a-b)^2\geq 0\implies a^3\geq 2a^2b-ab^2\implies \frac{a^3}{b}\geq2a^2-ab$$ and so $$\sum\frac{a^3}{b}\geq 2\sum a^2 - \sum ab\geq \sum ab.$$

Or you can directly use AM-GM by figuring out the coefficients by writing: $$x\frac{a^3}{b} + y\frac{b^3}{c}+z\frac{c^3}{a}\geq (x+y+z)\sqrt[x+y+z]{a^{3x-z}b^{3y-x}c^{3z-y}} = ab,$$ for example.

• I reckon $x=5/13$, $y=6/13$ and $z=2/13$ will do. Commented Aug 25, 2020 at 19:06
• nice method but i guess the starting point is to consider the expression $a({a-b)}^2$ how did you know that we should start with that Commented Aug 26, 2020 at 2:34
• @Quantum It's not known before. See my solution. Commented Aug 26, 2020 at 13:15

Another way.

By AM-GM: $$\sum_{cyc}\frac{a^3}{b}=\frac{1}{13}\sum_{cyc}\left(\frac{5a^3}{b}+\frac{6b^3}{c}+\frac{2c^3}{a}\right)\geq\sum_{cyc}\sqrt[13]{a^{15-2}b^{-5+18}c^{-6+6}}=\sum_{cyc}ab.$$

I wanted to present an elementary solution: $${a^3 \over b} + ab \ge 2a^2$$ and by $$2$$ other cyclic inequalities, $$\sum_{cyc}{a^3\over b}\geq 2(a^2+b^2+c^2)-ab-bc-ca\ge ab+bc+ac$$ Now it's sufficient, $$a^2+b^2+c^2\ge ab+bc+ca \Rightarrow (a-b)^2+(b-c)^2+(c-a)^2\ge 0$$

• well ,that was a clever way Commented Aug 26, 2020 at 15:17
• @Quantum See my solution to your latest problem. Commented Aug 26, 2020 at 15:19
• yes i saw that too it was nice i have already upvoted.thanks!! Commented Aug 26, 2020 at 15:20