Prove that $ \left(1+\frac a b \right) \left(1+\frac b c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$.

Hint: $$\frac{a}{b} + \frac{a}{c} + 1 \ge 3 \frac{a}{\sqrt[3]{abc}}$$ by AM-GM.


We can begin by clearing denominators as follows

$$a^2c+a^2b+b^2a+b^2c+c^2a+c^2b\geq 2a^{5/3}b^{2/3}c^{2/3}+2a^{2/3}b^{5/3}c^{2/3}+2a^{2/3}b^{2/3}c^{5/3}$$

Now by the Arithmetic Mean - Geometric Mean Inequality,

$$\frac{2a^2c+2a^2b+b^2a+c^2a}{6} \geq a^{5/3}b^{2/3}c^{2/3}$$

That is,

$$\frac{2}{3}a^2c+\frac{2}{3}a^2b+\frac{1}{3}b^2a+\frac{1}{3}c^2a \geq 2a^{5/3}b^{2/3}c^{2/3}$$

Similarly, we have

$$\frac{2}{3}b^2a+\frac{2}{3}b^2c+\frac{1}{3}c^2b+\frac{1}{3}a^2b \geq 2a^{2/3}b^{5/3}c^{2/3}$$ $$\frac{2}{3}c^2b+\frac{2}{3}c^2a+\frac{1}{3}a^2c+\frac{1}{3}b^2c \geq 2a^{2/3}b^{2/3}c^{5/3}$$

Summing these three inequalities together, we obtain the desired result.


The same inequality mentioned by @Sanchez used three times takes us to: $$\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+3\geq \frac{3(a+b+c)}{\sqrt[3]{abc}}$$ On the other hand, AM-GM gives us: $$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$$ so that: $$-3 \geq -\frac{a+b+c}{\sqrt[3]{abc}}$$ And adding the first and third inequalities gives the result.

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Inequality

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