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let $ABC$ be an acute triangle with all angles greater than $45^o$ Prove that $$\frac{2}{1+\tan A}+\frac{2}{1+\tan B}+\frac{2}{1+\tan C} \le 3(\sqrt{3}-1)$$

I let $\tan A=a$, $\tan B=b$, $\tan C=c$ with $a+b+c=abc$ then the inequality equivalent to $\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c} \le 3(\sqrt{3}-1)$ but it's not work.

I don't known what to do. Algebra or trigonometric?

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1  
Try: Put $C=180-(A+B)$ –  Rahul Taneja Aug 28 '12 at 15:20

3 Answers 3

up vote 3 down vote accepted

What about Calculus:

$f(x)=\frac{1}{1+\tan(x)}$ has a positive second derivative on $(\frac{\pi}{4}, \frac{\pi}{2})$. This solves the problem.

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Let's use Lagrange multiplier method to maximize $$ f(a,b,c) = \frac{2}{1+a} + \frac{2}{1+b} + \frac{2}{1+c} $$ subject to the constraint $0 = g(a,b,c) = a+b+c-a b c$. Derivatives of $f - \lambda g$ read: $$ \begin{eqnarray} \frac{\partial(f- \lambda g)}{\partial a} &=& -\frac{2}{(1+a)^2} + \lambda (b c -1) = \frac{-2+ \lambda (b c-1)(1+a)^2}{(1+a)^2} = 0\\ \frac{\partial(f- \lambda g)}{\partial b} &=& -\frac{2}{(1+b)^2} + \lambda (a c -1) = \frac{-2+ \lambda (a c-1)(1+b)^2}{(1+b)^2} = 0\\ \frac{\partial(f- \lambda g)}{\partial c} &=& -\frac{2}{(1+c)^2} + \lambda (a b -1) = \frac{-2+ \lambda (a b-1)(1+c)^2}{(1+c)^2} = 0 \end{eqnarray} $$ Eliminating out $\lambda$ this results into two equation: $$ (1+a)^2 (b c-1) = (1+b)^2 (a c-1) , \quad (1+b)^2 (a c-1) = (1+c)^2 (a b-1) $$ substituting $c = \frac{a+b}{a b-1}$ we get: $$ \frac{(1+a)^2 (1+b^2}{a b -1} = \frac{(1+a^2)(1+b)^2}{a b -1}, \quad \frac{(1+a^2)(1+b)^2}{a b -1} = \frac{\left(a b + a +b -1\right)^2}{a b -1} $$ Subtracting: $$ \frac{2 (b-a)(a b -1)}{a b-1} = 0, \quad \frac{4 b - 2 a(b^2-1)}{a b -1} = 0 $$ Giving $a=b$ and $a^2 = 3$. Thus $a=b=c=\sqrt{3}$ is the extremum. It is easy to see that this is a maximum. The inequality, we see, is saturated at $a=b=c=\sqrt{3}$, corresponding to $A=B=C=\frac{\pi}{3}$.

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Putting $D=A-45^{\circ}$ etc, $D+E+F=(A+B+C-135^{\circ})=45^{\circ}$

$1+\tan A=1+\tan(D+45^{\circ})=\frac{2}{1-\tan D}$ appyling $\tan(A+B)=\frac{\tan A+ \tan B}{1-\tan A \tan B}$

So, $\frac{2}{1 + \tan A}=1-\tan D$

Now the problem reduces to minimize $\tan D+\tan E+ \tan F$ where $D+E+F=45^{\circ}$ and $D,E,F>0$

If we can show that the minimum value occurs when $D=E=F$, we are done.

(i) Observe that $\tan x $ is convex for $x=D,E$ or $F$, as $\frac{d\tan x}{dx}=sec^2x$ and $\frac{d^2\tan x}{dx^2}=2\sec^2x\tan x>0$ as $D+E+F=45^{\circ}$ and $D,E,F>0$.

Using the Jensen's Inequality, we get that $\sum_{i=1}^n \tan A_i \ge n \cdot \tan \left( \frac{A_1 + A_2 + \cdots+ A_n}{n} \right) $ if $\sum_{i=1}^n A_i$ is constant.

So, the value of $\tan D+\tan E+ \tan F$ is minimized when $D=E=F$.

(ii) Alternatively, $\tan D+\tan E+ \tan F=\tan D+\tan E+ \tan (45^{\circ} - D - E)=g(D,E)$(say),

Applying partial derivative wrt to D, $\frac{\partial g}{\partial D}=\sec^2D+sec^2(45^{\circ} - D - E)\cdot (-1)$

$\frac{\partial g}{\partial D}=0=>\sec^2D=sec^2(45^{\circ} - D - E)=>D=±(45^{\circ} - D - E)$

If $D=-(45^{\circ} - D - E)$, then $E=45^{\circ}$ which is impossible as $D,E,F>0$,

so, $D=45^{\circ} - D - E=>E=45^{\circ} - 2D$

Putting this is in $D+E+F=45^{\circ}=>F=D$

Applying partial derivative wrt to E, $\frac{\partial g}{\partial E}=\sec^2E+sec^2(45^{\circ} - D - E)\cdot (-1)$

$\frac{\partial g}{\partial E}=0=>\sec^2E=sec^2(45^{\circ} - D - E)$

So, $\sec^2E=\sec^2D=>D=E$ as $D,E,F>0$,

So, $\frac{\partial g}{\partial E}=0=\frac{\partial g}{\partial D}=>D=E=F$

Using Second Derivative Test for $g(D,E)$, we can show that $min(g(D,E))=g(D,E)_{D=E=F}=g(15^{\circ}, 15^{\circ})$

So, the minimum value of $\tan D+\tan E+ \tan F$ is $3 (\tan15^{\circ})=3(2-\sqrt 3)$

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