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)$.

Solution 1:

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

Solution 2:

I know it is a very late to answer this, but i came across a very nice method to prove this. $$\displaystyle \left(1+\frac a b \right) \left(1+\frac b c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$$ $$\displaystyle \iff (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right) \geq 3+2\left(\frac{a+b+c}{\sqrt[3]{abc}}\right)$$ $$\displaystyle \iff \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right) \geq \frac{3}{a+b+c}+\frac{2}{\sqrt[3]{abc}}$$ Replacing by their inverses $$\iff 3AM\ge HM+2GM$$ Which is evident.

Solution 3:

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.