Show that : $\frac{a+b}{a+3b+2c}+\frac{b+c}{2a+b+3c}+\frac{a+c}{3a+2b+c} \geq 1$

Question

question

Show that for any strictly positive real numbers a, b, c the inequality holds:

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

and specify when the tie occurs.

my idea

After some examples I got to the conclusion that equality occurs only when $a=b=c$.

This fact got me thinking that equality should be proved using the inequality means, which equality occurs when all the elements are equal.

I tried using these inequalities but got nothing useful. Then I tried of trying CBS or the inequality of Bermstrong but again nothing useful.

The thing is that I always have this problem when demonstratng inequality: I don't know where to start.

Hope one fo you can help me! Thank you!

Answer 1

You claimed to CS it but got nothing useful. I would challenge you to retry it.

Here's a one-line naive CS solution via Titu's lema with the standard trick of making the numerator a perfect square:

$$ \sum \frac{ a+b}{a+3b+2c} = \sum \frac{ (a+b)^2} { (a+b)(a+3b+2c)} \\ \geq \frac{ 4(a+b+c)^2 }{ 4(a^2+b^2+c^2) + 8 (ab+bc+ca)} = 1. $$

It is left to the reader to check the equality conditions and verify that equality holds iff $a=b=c$.

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Note: I strongly reommend you to have a go at these.

1. The substitution of $a+b = x$ that NN2 indicated turns this into a well-known CS question. In fact, in my comment, I was trying to get you to show a harder (though equivalent) inequality that

$$ \sum \frac{x}{x+2y} \geq 1 \geq \sum \frac{y}{x+2y}. $$

2. Another trick would be to subtract $1/3$ from each term and show that

$$ \sum \frac{a}{a+3b+2c} \geq \sum \frac{c}{a+3b+2c}.$$