Let $a,b,c$ be positive real numbers. Prove that $$\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \geq 1$$

I was trying to solve the problem and I found some difficulties in my solution. Here is the solution:

From Cauchy-Schwarz, we have $$\frac{(\sqrt a)^2}{b+2c}+\frac{(\sqrt b)^2}{c+2a}+\frac{(\sqrt c)^2}{a+2b} \geq \frac{(\sqrt a +\sqrt b +\sqrt c)^2}{3(a+b+c)}.$$ Then we have to prove that $$(\sqrt a +\sqrt b +\sqrt c)^2 \geq 3(a+b+c)\\ \implies a+b+c+2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\geq 3(a+b+c)\\ \implies \sqrt{ab}+\sqrt{bc}+\sqrt{ca}\geq a+b+c$$ which is not true because $$(a+b+c)(b+c+a) \geq (\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^2\\ \implies a+b+c \geq\sqrt{ab}+\sqrt{bc}+\sqrt{ca}$$ by Cauchy-Schwarz.

I'm quite sure that the problem statement is correct. So, I made a mistake somewhere in my solution. But I am unable to find that. So, I basically want to know where my mistake is and I don't want alternative solution to the problem.

There is a similar problem here but that doesn't answer my question because I don't want the solution of the problem rather I want to know where the mistake is in my solution.

As @Macavity wrote, I used an inequality which is not tight enough. Then, my question is how to know if an inequality I'm using is tight enough or not.

(This issue is bothering me a lot. Today I was proving that $$\frac{a^2}{(a+b)(a+c)}+\frac{b^2}{(b+c)(b+a)}+\frac{c^2}{(c+a)(c+b)} \geq \frac 3 4$$ for all positive reals $a,b,c$ and the same thing happened as above. I really want to know how to avoid this.)

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    $\begingroup$ There is no mistake. You have just used an inequality which is not tight enough, so you got a lower bound which is not really the minimum. Note the implications you've used cannot be used in reverse direction, so you aren't contradicting anything either. Instead if you had used CS slightly differently, it would have been tight enough to get the desired minimum. $\endgroup$
    – Macavity
    Jul 7, 2021 at 11:50
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    $\begingroup$ Yes, that's a good way to interpret it. On how to ensure you're working with tight enough inequalities- there is no assured way i know. A necessary condition is to maintain points of equality throughout (ie here when $a=b=c$). In this particular case, applying CS on the form $\sum \frac{(\sqrt a)^2}{b+2c}$ is not good enough, though applying on $\sum \frac{a^2}{ab+2ca}$ is sufficient- a general rule seems difficult to state. $\endgroup$
    – Macavity
    Jul 7, 2021 at 16:14
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    $\begingroup$ The question of "how do I know if an inequality is tight enough?" is tantamount to asking "how do I know how to solve inequalities?" The answer is that you should do a lot of problems, and get used to the fact that sometimes a certain path to a solution won't work. $\endgroup$
    – user147556
    Jul 8, 2021 at 18:47
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    $\begingroup$ Try to multiply with $a$, $b$, $c$ for each term, then use the same inequality you used. $\endgroup$
    – N.Quy
    Jul 10, 2021 at 12:47
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    $\begingroup$ it's practice and practice. If you continue solving inequalities, at some point this exact issue you are having and then hopefully resolving will be used somewhere in the future when you do a similar problem. AM-GM, Cauchy-Schwarz are beginner methods and they almost always are futile when tackling tight inequalities. Stronger tools include Schur, Muirhead, Jensen's and some known inequalities like the Iran's inequality etc. The Holy Grail is of course expanding everything and use something like $uvw$ method or Vasc's Equal Variable Theorem. $\endgroup$
    – dezdichado
    Jul 10, 2021 at 23:03

1 Answer 1


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Here is my soluation. I used a cauchy bunyakovsky instead of cauchy schwarz.

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    $\begingroup$ Why can't you use MathJax? $\endgroup$ Jan 4 at 11:32

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