Proving an equality using given ones; no need for differentiation Prove that $(\frac ab+\frac bc+\frac ca)(\frac ba+\frac cb+\frac ac)\geqslant9$
The formulas given were
$$\frac{a+b} {2}\geqslant\sqrt {ab}$$
$$a^2+b^2\geqslant2ab$$
$$\frac{a+b+c} {3}\geqslant\root3\of{abc}$$
$$a^3+b^3+c^3\geqslant3abc$$
for all $a\gt0, b\gt0, c\gt0.$
I couldn't think of any way to do this; please help.
 A: Use that $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3\sqrt[3]{\frac{a}{b}\frac{b}{c}\frac{c}{a}}=3$$
and $$\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\geq 3\sqrt[3]{\frac{b}{a}\frac{c}{b}\frac{a}{c}}=3$$
A: Try and apply $x + y + z \geq 3 \sqrt[3]{xyz}$ to each factor in the left-hand side of the inequality to prove.
A: Method 1: AM-GM
Apply the last given inequality, but with the fractions instead.
$$\frac ab+\frac bc+\frac ca\ge 3 \sqrt[3]{\frac ab\,\frac bc\,\frac ca} = 3.$$
Do the same for $\frac ba+\frac cb+\frac ac$
Method 2: Cauchy Schwarz
$$9 = 3^2 = \left(\frac{\sqrt{a}}{\sqrt{b}} \, \frac{\sqrt{b}}{\sqrt{a}} + \frac{\sqrt{b}}{\sqrt{c}} \, \frac{\sqrt{c}}{\sqrt{b}} + \frac{\sqrt{c}}{\sqrt{a}} \, \frac{\sqrt{a}}{\sqrt{c}}\right)^2 \\ \le \left(\frac ab + \frac bc + \frac ca \right) \left(\frac ba + \frac cb + \frac ac \right)$$
A: Also, we can use the first formula:
$$\prod_{cyc}\frac{a}{b}\prod_{cyc}\frac{a}{c}=\sum_{cyc}\left(\frac{a^2}{bc}+\frac{bc}{a^2}+1\right)\geq\sum_{cyc}\left(2\sqrt{\frac{a^2}{bc}\cdot\frac{bc}{a^2}}+1\right)=9.$$
