# Prove $\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab} \ge 1$

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

-
Thank you, I will remember. – harrypham Jan 19 '13 at 13:49

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,

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

-

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.

-
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.

-

multiply abc

$a+b+c\geq abc$

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

$a^2+b^2+c^2-(a+b+c)$

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

therefore

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

-