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If $A,B>0$ and $\displaystyle A+B = \frac{\pi}{3},$ Then find Maximum value of $\tan A\cdot \tan B$.

$\bf{My\; Try::}$ Given $$\displaystyle A+ B = \frac{\pi}{3}$$ and $A,B>0$.

So we can say $$\displaystyle 0< A,B<\frac{\pi}{3}$$. Now taking $\tan $ on both side, we get

$$\displaystyle \tan(A+B) = \tan \left(\frac{\pi}{3}\right).$$ So $\displaystyle \frac{\tan A+\tan B}{1-\tan A\tan B} = \sqrt{3}$.

Now Let $\displaystyle \tan A\cdot \tan B=y\;,$ Then $\displaystyle \tan B = \frac{y}{\tan A}.$

So $$\displaystyle \frac{\tan A+\frac{y}{\tan A}}{1-y}=\sqrt{3}\Rightarrow \tan^2 A+y=\sqrt{3}\tan A-y\sqrt{3}\tan A$$

So equation $$\tan^2 A+\sqrt{3}\left(y-1\right)\tan A+y=0$$

Now for real values of $y\;,$ Given equation has real roots. So its $\bf{Discrimnant>0}$

So $$\displaystyle 3\left(y-1\right)^2-4y\geq 0\Rightarrow 3y^2+3-6y-4y\geq 0$$

So we get $$3y^2-10y+3\geq 0\Rightarrow \displaystyle 3y^2-9y-y+3\geq 0$$

So we get $$\displaystyle y\leq \frac{1}{3}\cup y\geq 3$$, But above we get $\displaystyle 0<A,B<\frac{\pi}{3}$

So We Get $$\bf{\displaystyle y_{Max.} = \left(\tan A \cdot \tan B\right)_{Max} = \frac{1}{3}}.$$

My Question is can we solve above question using $\bf{A.M\geq G.M}$ Inequality or Power Mean equality.

Thanks

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3 Answers 3

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$$(\tan A+\tan B)^2-4\tan A\tan B=(\tan A-\tan B)^2\ge0$$ $$\iff4\tan A\tan B\le(\tan A+\tan B)^2$$

Also by $\bf{A.M\geq G.M},$ $\dfrac{\tan A+\tan B}2\ge\sqrt{\tan A\tan B}$

By Jensen's inequality, $$\frac{\tan A+\tan B}2\le\tan\frac{A+B}2=\frac1{\sqrt3}$$

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By using $1-\tan A\tan B=\frac{\tan A+\tan B}{\tan(A+B)}$, it follows that for $A>0,B>0,A+B=\pi/3$, $$\tan A\tan B=\frac{\tan A(\sqrt{3}-\tan A)}{1+\sqrt{3}\tan A}.$$ Hence $$\begin{align}\max_{A>0,B>0,A+B=\pi/3} \tan A\tan B&=\max_{0<A<\pi/3} \frac{\tan A(\sqrt{3}-\tan A)}{1+\sqrt{3}\tan A}=\max_{1<s<4} \frac{(s-1)(4-s)}{3s}\\ &=\frac{1}{3}\max_{1<s<4}\left(5-\left(s+\frac{4}{s}\right)\right)= \frac{1}{3}\left(5-\min_{1<s<4}\left(s+\frac{4}{s}\right)\right) \\&=\frac{1}{3}(5-4)=\frac{1}{3}\end{align}$$ where $s=1+\sqrt{3}\tan A$ and by AGM-inequality $$s+\frac{4}{s}\geq 2\sqrt{s\cdot \frac{4}{s}}=4$$ with equality for $s=2\in(1,4)$, that is $A=B=\pi/6$.

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$$ \tan A\tan B=\tan A\tan\big(\frac{\pi}{3}-A\big)=\tan A.\frac{\sqrt{3}-\tan A}{1+\sqrt{3}\tan A}=1-1+\frac{\sqrt{3}\tan A-\tan^2A}{1+\sqrt{3}\tan A}\\ =1+\frac{-1-\sqrt{3}\tan A+\sqrt{3}\tan A-\tan^2A}{1+\sqrt{3}\tan A}=1-\frac{1+\tan^2A}{1+\sqrt{3}\tan A}=1-\frac{\sec^2A}{1+\sqrt{3}\tan A}\\ =1-\frac{2}{2\cos^2A+\sqrt{3}.2\sin A\cos A}=1-\frac{2}{1+2\big[\frac{1}{2}\cos 2A+\frac{\sqrt{3}}{2}\sin 2A\big]}\\ =1-\frac{2}{1+2\sin\big(\frac{\pi}{6}+2A\big)} $$ $\tan A\tan B$ is maximum $\implies \frac{2}{1+2\sin\big(\frac{\pi}{6}+2A\big)}$ is minimum $\implies \sin\big(\frac{\pi}{6}+2A\big)$ is maximum$\implies \sin\big(\frac{\pi}{6}+2A\big)=1$ $$ (\tan A\tan B)_\text{max}=1-\frac{2}{1+2.1}=1-\frac{2}{3}=\frac{1}{3} $$

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  • $\begingroup$ Did you ask the same question at math.stackexchange.com/questions/2958959/… ? $\endgroup$
    – Robert Z
    Commented Oct 17, 2018 at 9:51
  • $\begingroup$ @RobertZ thats right. when I posted i could not find this post. i thought its worth posting this answer in the original post. $\endgroup$
    – Sooraj S
    Commented Oct 17, 2018 at 10:11

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