Prove that for positive real number $a,b,c$ such that $abc=1$ the following Inequality holds

inequality

Prove that for positive real number $a,b,c$ such that $abc=1$ the following Inequality holds: $$ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{a+b+2} + \frac{1}{b+c+2} + \frac{1}{c+a+2} \geq \frac{15}{4} $$

Letting $a= \frac{x}{y},b=\frac{y}{z}, c=\frac{z}{x}$, we get the Inequality is equivalent to
$$\frac{y}{x}+\frac{z}{y}+\frac{x}{z}+\frac{1}{\frac{x}{y}+\frac{y}{z}+2}+\frac{1}{\frac{y}{z}+\frac{z}{x}+2}+\frac{1}{\frac{x}{y}+\frac{z}{x}+2} \geq \frac{15}{4}$$. I tried some algebraic manipulations but those didn't work. I want some instructive hints to make progress. Thank you

Best Answer

If you're stuck, explain what you've tried with each of these hints, and I can figure out a further hint.

Hint 1: Apply CS (Titu's lemma) on $ \sum \frac{1}{ a+b+2 } $ to get a lower bound that depends only on $S =a+b+c$.

$ \sum \frac{1}{ a+b+2} \geq \frac{ 9}{ 2(a+b+c) + 6 } $.

Because $ \sum \frac{1}{a+b+2} \leq \frac{3}{4}$ (see note below), so we need a way to bound this term from below.
Titu's Lemma is a good first stab for "when we have got 1 [or anything] on numerator and a monstrous expression on the denominator [multiply by anything as needed]", especially if the denominator is cyclic and combining the terms makes it "much nicer".

Hint 2: For $abc=1$, $ S = a+b+c \geq 3$.

This should be obvious / well-known, but it motivates the next hint. See if you can deduce Hint 3 without revealing it.

Hint 3:

For $ abc = 1$, show that $\sum \frac{1}{a} \geq \sqrt{3(a+b+c) } = \sqrt{3S}$.

Finally, putting this all together, show that:

For $ S \geq 3$, $ \sqrt{3S} + \frac{9}{2S + 6 } \geq \frac{15}{4}$.
Equality holds iff $S = 3$, which means $ a = b = c = 1$.

If you want to avoid calculus (which was my gut approach given the simplicity), show that $$ \sqrt{3S} + \frac{9}{2S + 6 } \geq \sqrt{3S} + \frac{ 9 } { 4 \sqrt{3S} } \geq \frac{15}{4}.$$

Note:

  • To show that $ \sum \frac{1}{a+b+2} \leq \frac{3}{4}$, use OP's substitution and work with $ \sum \frac{z}{x+y+2z} \leq \frac{3}{4} \Leftrightarrow \frac{x+y}{x+y+2z} \geq \frac{3}{2}$, which is true by Titu's lemma (applied in a direct manner by multiplying the numerator to the denominator).
  • This tells us that we can't simply hope to show separately that $ \sum \frac{1}{a} \geq 3$ (which is true) and $ \sum \frac{1}{a+b+2} \geq \frac{3}{4} $(which is untrue). We need to somehow combine both parts together, using the surplus in the first to offset the deficit in the second.
Related Question