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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!

IONELA BUCIU
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  • Some of the "standard" CS tricks work well on this. $\quad$For a similar question, can you prove that $\sum x/(x+2y) \geq \sum y/(x + 2y) $ using CSB? If so, please edit that in. – Calvin Lin Jan 23 '24 at 08:45

2 Answers2

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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 $a=b=c$.

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

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

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

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

Calvin Lin
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Denote $(x,y,z) = (a+b,b+c,c+a)$ then $$\begin{align} &\iff \frac{x}{x+2y}+\frac{y}{y+2z}+\frac{z}{z+2x} \ge 1 \\ &\iff \frac{4 x^2 y + 4 x y^2 + 4 x^2 z + 3 x y z + 4 y^2 z + 4 x z^2 + 4 y z^2}{(x + 2 y) (2 x + z) (y + 2 z)} \ge 1 \\ &\iff 4 x^2 y + 4 x y^2 + 4 x^2 z + 3 x y z + 4 y^2 z + 4 x z^2 + 4 y z^2 \ge x + 2 y) (2 x + z) (y + 2 z)\\ &\iff 2 (x^2 y + y^2 z + x z^2- 3 x y z ) \ge 0 \tag{1} \end{align}$$ $(1)$ holds true by AM-GM inequality.

The equality occurs if and only if $x = y = z \iff a = b=c$.

NN2
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