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For $0 \lt a, b, c \lt 1$, if $ab + bc + ca = 1$, show that $$\frac a{1 - a^2} + \frac b{1 - b^2} + \frac c{1 - c^2} \ge \frac {3 \sqrt 3}2.$$

I want to use trigonometric substitution:

For the angles $A, B, C$ of any acute triangle, $$\tan A + \tan B + \tan C = \tan A \tan B \tan C,$$ $$\frac 1{\tan A \tan B} + \frac 1{\tan B \tan C} + \frac 1{\tan C \tan A} = 1.$$

Also, $\tan A, \tan B, \tan C \gt 0$. So I substitute $a, b, c$ for $\frac 1{\tan A}, \frac 1{\tan B}, \frac 1{\tan C}$ respectively. Then the inequality in question becomes $$\frac {\tan A}{1 - \tan^2 A} + \frac {\tan B}{1 - \tan^2 B} + \frac {\tan C}{1 - \tan^2 C} \le -\frac {3 \sqrt 3}2.$$

Here $A, B, C \not = \frac {\pi}4$ since $a, b, c \not = 1$.

By the trigonometric identity $\tan 2A = \frac {2 \tan A}{1 - \tan^2 A}$, we have

$$\tan 2A + \tan 2B + \tan 2C \le -3 \sqrt 3,$$

where $0 \lt A, B, C \lt \frac {\pi}2$, $A, B, C \not = \frac {\pi}4$, and $A + B + C = \pi$.

How do I proceed?

Edit: The restriction $a, b, c \lt 1$ was added after the question had received some answers, thanks to Michael Rozenberg, who pointed out this mistake.

Yuxiao Xie
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4 Answers4

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hint: Lets use $a = \tan\left(\frac{A}{2}\right)$. You can define $b, c$ similarly, then your inequality becomes: $\tan(A) + \tan(B) + \tan(C) \geq 3\sqrt{3}$, with $0 < A,B,C < \dfrac{\pi}{2}$ and $A+B+C = \pi$. And this inequality is standard result of convexity of $\tan(x)$ over $\left(0,\frac{\pi}{2}\right)$.

DeepSea
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It's obviously wrong! Try $a=1.01$ and $b=c=\sqrt{2.0201}-1.01$.

Michael Rozenberg
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    Since you are giving an answer that disagrees with both the Question and the two other posted Answers, you might help your Readers follow your objection more easily by spelling out the computation. – hardmath Apr 10 '16 at 21:25
  • @hardmath $\sum\limits_{cyc}\frac{a}{1-a^2}-\frac{3\sqrt3}{2}=-51.8567...$ for my counterexample. – Michael Rozenberg Apr 10 '16 at 22:32
  • Okay, my arithmetic (I made a Google Sheet) gives $-49.25865721$ for the expression that is supposed to be bounded below, but the sum $ab+bc+ca$ does come out to be $1$. So I agree with the basic conclusion. – hardmath Apr 10 '16 at 22:42
  • @hardmath I think your are wrong with $-49.25...$ – Michael Rozenberg Apr 11 '16 at 02:54
  • @hardmath I wrote $\frac{3\sqrt3}{2}$ before and if you got another number so it's your mistake and it's not my mistake. – Michael Rozenberg Apr 11 '16 at 03:08
  • Put it this way: Since I got $\frac{a}{1-a^2} + \frac{b}{1-b^2} + \frac{c}{1-c^2} = −49.25865721$, this is the expression that is supposed to be bounded below by $\frac{3\sqrt{3}}{2}$. If we now subtract $\frac{3\sqrt{3}}{2}$, we get $−49.25865721 - 2.59807621135 = -51.8567334214$. It matches your figure, just as you wrote the subtraction. – hardmath Apr 11 '16 at 03:37
  • Note that there was a comment, now removed, by someone else who thought that $ab + bc + ca = 1$ was not true with your example. The point I was making there is that it is true. – hardmath Apr 11 '16 at 03:41
  • Maybe I missed some restrictions to $a, b, c$. And since you've provided your counterexample, the proofs provided ought to be wrong. Can you point out where the wrong step is and how can I refine my question? – Yuxiao Xie Apr 11 '16 at 04:43
  • Maybe adding $a, b, c \lt 1$ will do? Okay, I just noticed that in the trigonometric substitution it is assumed that $a, b, c \lt 1$, since $a = \tan \frac A2 \lt \tan \frac {\pi}4 = 1$, where $0 \lt A \lt \frac {\pi}2$. I'll edit it. Thank you @MichaelRozenberg. But actually I think this should posted as a comment instead of an answer... – Yuxiao Xie Apr 11 '16 at 04:56
  • @YuxiaoXie: That is the restriction I suspect is needed (so $a,b,c \in (0,1)$). But I would defend Michael's posting this as an Answer, because it does settle your original framing of the Question (I have no problem with your adjusting the problem to take account of Michael's observation). – hardmath Apr 11 '16 at 13:36
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We can prove $S=\sum\tan A=\prod\tan A$

Using AM GM inequality if $\tan A,\tan B,\tan C\ge0,$ $$\dfrac{\sum\tan A}3\ge\sqrt[3]{\prod\tan A}$$

$$\iff\dfrac S3\ge\sqrt[3]S\implies\left(\dfrac S3\right)^3\ge S\iff S^2\ge27$$ as $S\ne0$

Can you take it from here?

lab bhattacharjee
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If you only want to prove it with AM-GM, I will give one proof using AM-GM and Cauchy-Schwarz. Note that the given condition implies that $a+b+c\geq\sqrt 3$. In addition, I will use the following two inequalities. $$(a+b+c)(ab+bc+ca)\leq a^3+b^3+c^3+6abc$$ and $$abc\leq\big(\dfrac{ab+bc+ca}{3}\big)^{\frac{3}{2}}.$$ Now, $$\sum\limits_{a,b,c} \dfrac{a}{1-a^2}\sum\limits_{a,b,c} a(1-a^2)\geq (a+b+c)^2.$$ But $\sum\limits_{a,b,c} a(1-a^2) = (a+b+c)(ab+bc+ca)-a^3-b^3-c^3\leq abc\leq \dfrac{6}{3\sqrt 3}=\dfrac{2}{\sqrt 3}$. These two combined together implies that $$\sum\limits_{a,b,c} \dfrac{a}{1-a^2}\geq\dfrac{(a+b+c)^2}{\dfrac{2}{\sqrt 3}}\geq\dfrac{3 \sqrt 3}{2}.$$
The first inequality is called Schur's inequality.

EDIT.

Actually an even easier solution is obtained by using $$\dfrac{a}{1-a^2}+\dfrac{9}{4}a(1-a^2)\geq 3a,$$ by AM-GM. Thus, $\sum\limits_{a,b,c} \dfrac{a}{1-a^2}\geq\dfrac{3}{4}\sum\limits_{a,b,c} a+\dfrac{9}{4}\sum\limits_{a,b,c} a^3\geq\dfrac{3}{4}\cdot\sqrt 3 + \dfrac{9}{4}\cdot\dfrac{1}{\sqrt 3} = \dfrac{3 \sqrt 3}{2}$. That $a+b+c\geq\sqrt 3$ and $a^3+b^3+c^3\geq\dfrac{1}{\sqrt 3}$ are trivial from the given condition.

dezdichado
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