Problem manipulating AM-GM with 3 variables

Question

For $a, b, c > 0$, prove $$ \frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \ge ab + bc + ca. $$

I am not sure whether I should cube the 3-variable AM-GM inequality.

Answer 1

Are you set on using AM-GM? It's pretty easy with Engel's form. We get $$\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \ge \frac{(a^2+b^2+c^2)^2}{ab+bc+ca}\ge ab+bc +ca.$$ The latter inequality is true because it is equivalent to $$(a-b)^2+(b-c)^2+(c-a)^2\ge 0.$$

Answer 2

\begin{eqnarray*} ac(a+b)(a-b)^2+ba(b+c)(b-c)^2+cb(c+a)(c-a)^2 \geq 0 \end{eqnarray*} and rearrange.

Answer 3

I don't think you need $AM\ge GM$.

Well by assume WLOG $a \ge b \ge c$ and thus $1/c \ge1/b \ge 1/a$ , by rearrangement (inequality) $$\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \ge\frac{a^3}{a}+\frac{b^3}{b}+\frac{c^3}{c} =a^2+b^2+c^2 \ge ab+bc+ac $$

Answer 4

By Cauchy-Schwartz inequality: $$\frac{P^2}{u}+\frac{Q^2}{v}+\frac{R^2}{w} \ge \frac{(P+Q+R)^2}{u+v+w}. u,v,w >0$$ So we have $$F=\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \ge \frac{(a^{3/2}+b^{3/2}+c^{3/2})^2}{a+b+c}.$$ Use Mean-Power-ineq.: $$\frac{(a^{3/2}+b^{3/2}+c^{3/2})}{3} \ge \left(\frac{a+b+c}{3}\right)^{3/2}$$ We get $$F \ge \frac{(a+b+c)^2}{3}\ge ab+bc+ca$$ Lastly we have used $$a^2+b^2+c^2 \ge ab+bc+ca$$

Answer 5

Multiply by 13. Use AM GM three times. $$5\frac{a^3}{b} + 6\frac{b^3}{c} + 2\frac{c^3}{a} \geq 13 \sqrt[13]{a^{13}b^{13}} = 13ab$$

If you have any questions I can answer :). Why I used 13 uses a simple trick, I can describe it to you.