Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $a,b,c$ be positive real numbers such that $\dfrac{1}{bc}+ \dfrac{1}{ca}+ \dfrac{1}{ab} \ge 1$. Prove that $\dfrac{a}{bc}+ \dfrac{b}{ca}+ \dfrac{c}{ab} \ge 1$.

share|cite|improve this question
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Julian Kuelshammer Jan 19 '13 at 13:34
Thank you, I will remember. – harrypham Jan 19 '13 at 13:49
up vote 3 down vote accepted

Denote $X=\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}$ and $Y=\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}$. In fact, if $X\ge 1$, then $Y\ge \sqrt{3}$. ($\sqrt{3}$ is optimal, because when $a=b=c=\frac{1}{\sqrt{3}}$, $X=1$ and $Y=\sqrt{3}$.)

By Cauchy-Schwarz inequality,

$$3(a^2+b^2+c^2)\ge (a+b+c)^2.$$

It follows that

$$Y=\frac{a^2+b^2+c^2}{abc}\ge \frac{(a+b+c)^2}{3abc}=\frac{abc}{3} X^2\ge \frac{abc}{3}.$$

If $abc\ge 3\sqrt{3}$, we are done. Otherwise, by inequality of arithmetic and geometric means,

$$Y\ge 3(abc)^{-\frac{1}{3}}\ge \sqrt{3}.$$

share|cite|improve this answer

Remember, that for any nonzero $x$, one has

$$ x+\frac{1}{x} \geq 2 \tag{1} $$

We apply this inequality repeatedly. Put

$$ T=\dfrac{a}{bc}+ \dfrac{b}{ca}+ \dfrac{c}{ab} $$

We have $$ T=\frac{a^2+b^2}{abc}+\frac{c}{ab} =\frac{\sqrt{a^2+b^2}}{ab} \bigg(\frac{\sqrt{a^2+b^2}}{c}+\frac{c}{\sqrt{a^2+b^2} }\bigg) \geq \frac{2(\sqrt{a^2+b^2})}{ab}, $$

by using (1) with $x=\frac{\sqrt{a^2+b^2}}{c}$. Then,

$$ T \geq \frac{2}{\sqrt{ab}} \sqrt{\frac{a}{b}+\frac{b}{a}} \geq \frac{2\sqrt{2}}{\sqrt{ab}} $$ by using (1) with $x=\frac{a}{b}$. So $T \geq \sqrt{\frac{8}{ab}}$.

We see that if $ab \leq 8$, we are done. So we may assume $ab \geq 8$. By symmetry, we may also assume $ac \geq 8, bc \geq 8$. But then

$$ 1 \leq \dfrac{1}{bc}+ \dfrac{1}{ca}+ \dfrac{1}{ab} \leq \frac{1}{8}+\frac{1}{8}+\frac{1}{8} =\frac{3}{8}, $$

which is impossible.

share|cite|improve this answer
I think you forgot a $\sqrt$ in the line after the "We have", in the numerator of the first factor in the third expression from the left. – user3533 Jan 19 '13 at 14:00
@user3533 : Fixed, thanks. – Ewan Delanoy Jan 19 '13 at 14:02
Very good answer. I don't understand the downvote. – user3533 Jan 19 '13 at 14:06
How do you justify the fact that assuming $ab\geq 8$ allows to assume $bc\geq 8$ and $ca\geq 8$? If it's boring to type, I would appreciate onilne links instead. P.S. it was not me who downvoted. – 007resu Jan 19 '13 at 14:09
@Freddy: If you take the proof that $ab \leq 8$ is enough and exchange $a$ and $c$ whenever they appear you will get a proof that $ca \leq 8$ is enough. – user3533 Jan 19 '13 at 14:15

I want to give a complete answer.

Assume by the sake of contradiction that $$\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}<1.\tag{1}$$ By Cauchy Schwarz it follows from $(1)$ that $$\frac{3}{abc}>\frac{3}{abc}\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\geq \left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)^2.\tag{2}$$

From $(2)$ we recover two facts

  • By $AM-GM$ mean $$\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)^2\geq 3\left(\frac{1}{a^2bc}+\frac{1}{ab^2c}+\frac{1}{abc^2}\right),$$ therefore, combined with $(2)$, we obtain on one hand $$1>\frac{1}{a}+\frac{1}{b}+\frac{1}{c},\tag{3}$$ which implies that $\min\{a,b,c\}>1$ (remember that they are all positive).
  • On the other hand we can also derive from $(2)$ and $(3)$ the following: $$\frac{1}{\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)}>\frac{a+b+c}{3}>1,\tag{4}$$

which in the end leads to an absurd with respect to the hypothesis of the problem.

Then $(1)$ must be false and the problem is solved.

share|cite|improve this answer

multiply abc

$a+b+c\geq abc $

$a^2+b^2+c^2\geq abc $


$⇔(a-\frac12)^2+(b-\frac12)^2+(c-\frac12)^2\geq \frac34$


$a^2+b^2+c^2\geq \frac34+abc$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.