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Question Number 15267 by Tinkutara last updated on 09/Jun/17

$$\mathrm{In}\mathrm{an}\mathrm{acute}\mathrm{angled}\mathrm{\Delta }ABC,\mathrm{the}$$$$\mathrm{minimum}\mathrm{value}\mathrm{of}\tan A\tan B\tan C\mathrm{is}?$$

Commented by prakash jain last updated on 09/Jun/17

$$A=0.1°,B=89.5°,C=89.5°$$$$\tan A\tan B\tan C=.518$$$$A=0.0000000001°$$$$B=C=89.99999999995°$$$$\tan A\tan B\tan C=4\times {10}^{-10}$$$$\mathrm{u}\mathrm{can}\mathrm{tanAtanBtan}\mathrm{C}asclose$$$$to0uwantminimumis0.$$$$\mathrm{These}\mathrm{calculated}\mathrm{values}\mathrm{are}\mathrm{wrong}\mathrm{as}$$$$\mathrm{highlighted}\mathrm{in}\mathrm{mrW}1\mathrm{comment}\mathrm{below}.$$

Commented by mrW1 last updated on 09/Jun/17

$$\mathrm{please}\mathrm{check}\mathrm{sir}:$$$$A=1°,B=89.5°,C=89.5°$$$$\tan A\tan B\tan C=229$$$$$$$$A=0.1°,B=89.95°,C=89.95°$$$$\tan A\tan B\tan C=2292$$$$$$$$A=0.001°,B=89.9995°,C=89.9995°$$$$\tan A\tan B\tan C=229189$$$$$$$$\tan A\tan B\tan C\mathrm{can}\mathrm{be}\mathrm{so}\mathrm{large}\mathrm{as}$$$$\mathrm{you}\mathrm{want},\mathrm{so}\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{maximum}.$$$$$$$$\mathrm{y}=\tan \mathrm{x}{\tan }^2(90-\frac{\mathrm{x}}{2})=\frac{\tan \mathrm{x}}{{\tan }^2\frac{\mathrm{x}}{2}}$$$$=\frac{2\tan \frac{\mathrm{x}}{2}}{{\tan }^2\frac{\mathrm{x}}{2}(1-{\tan }^2\frac{\mathrm{x}}{2})}$$$$=\frac{2}{\tan \frac{\mathrm{x}}{2}(1-\tan \frac{\mathrm{x}}{2})(1+\tan \frac{\mathrm{x}}{2})}$$$$\underset{x\to 0}{\lim }\mathrm{y}=2\times \underset{x\to 0}{\lim }\frac{1}{\tan \frac{\mathrm{x}}{2}}=\mathrm{\infty }$$

Commented by Tinkutara last updated on 09/Jun/17

$$\mathrm{But}\mathrm{answer}\mathrm{is}3\sqrt{3}.\mathrm{Maybe}\mathrm{its}\mathrm{by}$$$$\mathrm{assuming}\mathrm{\angle }A=\mathrm{\angle }B=\mathrm{\angle }C=\frac{\pi }{3}.\mathrm{But}$$$$\mathrm{how}?$$

Commented by prakash jain last updated on 09/Jun/17

$$\mathrm{Thanks}\mathrm{mrW}1.\mathrm{My}\mathrm{calculator}$$$$\mathrm{was}\mathrm{set}\mathrm{to}\mathrm{radians}\mathrm{when}\mathrm{i}\mathrm{did}$$$$\mathrm{the}\mathrm{above}\mathrm{calculation}.$$

Commented by prakash jain last updated on 10/Jun/17

$$\tan A\tan B\tan C=\tan A\tan B\tan (\pi -(A+B)$$$$-\tan A\tan B\frac{\tan A+\tan B}{1-\tan A\tan B}$$$$\tan A=x$$$$\tan B=y$$$$f(x,y)=-xy\frac{x+y}{1-xy}=-\frac{x^{2}y+{xy}^2}{1-xy}$$$$\frac{\partial f}{\partial x}=-\frac{(2xy+y^2)(1-xy)-(-y)(x^{2}y+{xy}^{2)}}{{(1-xy)}^2}$$$$=-\frac{2xy+y^2-2x^2{y}^2-{xy}^3+x^2{y}^2+{xy}^3}{{(1-xy)}^2}$$$$=-\frac{2xy+y^2-x^2{y}^2}{{(1-xy)}^2}$$$$\frac{\partial f}{\partial x}=0\Rightarrow 2xy+y^2-x^2{y}^2=0\left(i\right)$$$$\frac{\partial f}{\partial y}=0\Rightarrow 2xy+x^2-x^2{y}^2=0\left(ii\right)$$$$stationarypointwhen\frac{\partial f}{\partial x}=0\wedge \frac{\partial f}{\partial y}=0$$$$\left(i\right)-\left(ii\right)=0\Rightarrow y^2-x^2=0\Rightarrow x=\pm y$$$$foracuteanglex>0\wedge y>0\Rightarrow x=y$$$$putx=yin\left(i\right)$$$$2x^2+x^2-x^4=0$$$$3x^2-x^4=0\Rightarrow x=0\vee x=\sqrt{3},y=\sqrt{3}$$$$tocheckifx=\sqrt{3}andy=\sqrt{3}\mathrm{is}$$$$\mathrm{indeed}\mathrm{a}\mathrm{minima}\mathrm{and}\mathrm{not}\mathrm{a}\mathrm{saddle}$$$$\mathrm{point}\mathrm{or}\mathrm{maxima}\mathrm{following}$$$$\mathrm{condition}\mathrm{must}\mathrm{be}\mathrm{checked}atpoint$$$$(\sqrt{3},\sqrt{3})$$$$f_{xx}{f}_{yy}-f_{xy}^2>0$$$$f_{xx}>0,f_{yy}>0$$

Commented by mrW1 last updated on 09/Jun/17

$$\mathrm{this}\mathrm{is}\mathrm{the}\mathrm{exact}\mathrm{proof}.\mathrm{thanks}.$$

Answered by mrW1 last updated on 09/Jun/17

$$\mathrm{Due}\mathrm{to}\mathrm{the}\mathrm{symmetry}\mathrm{the}\mathrm{minimum}$$$$\mathrm{occurs}\mathrm{when}\mathrm{all}\mathrm{variables}\mathrm{are}\mathrm{equal},$$$$\mathrm{i}.\mathrm{e}.\mathrm{A}=\mathrm{B}=\mathrm{C}=60°$$$$\tan \mathrm{A}\tan \mathrm{B}\tan \mathrm{C}=\sqrt{3}\times \sqrt{3}\times \sqrt{3}=3\sqrt{3}$$$$$$$$\mathrm{This}\mathrm{can}\mathrm{be}\mathrm{proved}\mathrm{as}\mathrm{following}.$$$$\mathrm{To}\mathrm{make}\mathrm{the}\mathrm{thing}\mathrm{simple}\mathrm{let}\mathrm{us}\mathrm{keep}$$$$\mathrm{C}\mathrm{unchanged}\mathrm{and}\mathrm{see}\mathrm{what}\mathrm{happens}$$$$\mathrm{if}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{different}\mathrm{from}60°.$$$$$$$$\mathrm{When}\mathrm{A}=\mathrm{B}=60°,\mathrm{we}\mathrm{have}$$$$\tan \mathrm{A}\tan \mathrm{B}=\sqrt{3}\times \sqrt{3}=3$$$$$$$$\mathrm{Whenn}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{differ}\mathrm{from}60°,{\mathrm{let}}^{\prime }\mathrm{s}$$$$\mathrm{say}\mathrm{A}=60°+\delta \mathrm{and}\mathrm{consequently}$$$$\mathrm{B}=69°-\delta ,\mathrm{then}$$$$\tan \mathrm{A}\tan \mathrm{B}=\tan (60+\delta )\tan (60-\delta )$$$$=\frac{\tan 60+\tan \delta }{1-\tan 60\tan \delta }\times \frac{\tan 60-\tan \delta }{1+\tan 60\tan \delta }$$$$=\frac{{\tan }^{2}60-{\tan }^2\delta }{1-{\tan }^{2}60{\tan }^2\delta }$$$$=\frac{3-{\tan }^2\delta }{1-3{\tan }^2\delta }=\frac{3-{9\tan }^2\delta +{8\tan }^2\delta }{1-{3\tan }^2\delta }$$$$=3+\frac{{8\tan }^2\delta }{1-{3\tan }^2\delta }$$$$\because 0<\delta <30°\left(\mathrm{acute}\mathrm{triangle}\right)$$$$0<\tan \delta <\frac{1}{\sqrt{3}}$$$$0<{\tan }^2\delta <\frac{1}{3}$$$$0<{3\tan }^2\delta <1$$$$\therefore 1-{3\tan }^2\delta >0$$$$\Rightarrow 3+\frac{{8\tan }^2\delta }{1-{3\tan }^2\delta }>3=\tan 60°\tan 60°$$$$\mathrm{that}\mathrm{means}\mathrm{when}\mathrm{A},\mathrm{B}\left(\mathrm{and}\mathrm{also}\mathrm{C}\right)$$$$\mathrm{differ}\mathrm{from}60°,\mathrm{the}\mathrm{value}\mathrm{of}$$$$\tan \mathrm{A}\tan \mathrm{B}\tan \mathrm{C}\mathrm{will}\mathrm{increase},$$$$\mathrm{therefore}\mathrm{we}\mathrm{have}\mathrm{its}\mathrm{minimum}\mathrm{at}$$$$\mathrm{A}=\mathrm{B}=\mathrm{C}=60°,\mathrm{and}\mathrm{the}\mathrm{minimum}$$$$\mathrm{value}\mathrm{is}3\sqrt{3.}$$

Commented by Tinkutara last updated on 10/Jun/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

Answered by Tinkutara last updated on 12/Jun/17

$$\mathrm{I}\mathrm{found}\mathrm{the}\mathrm{best}\mathrm{method}:$$$$\mathrm{In}\mathrm{any}\mathrm{\Delta }ABC,\mathrm{we}\mathrm{have}$$$$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$$$=x\left(\mathrm{say}\right)$$$$\mathrm{Now}\frac{x}{3}⩾\sqrt[3]{x}[\because \mathrm{AM}⩾\mathrm{GM}]$$$$\frac{x^3}{27}⩾x$$$$x⩾3\sqrt{3}$$

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