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Question Number 207361 by hardmath last updated on 12/May/24

$$\mathrm{y}=\frac{\mathrm{tg}\mathbf{x}+\mathrm{ctg}\mathbf{x}}{8},(0;\frac{\pi }{2})$$$$\mathrm{Find}:\min \left(\mathrm{y}\right)=?$$

Answered by Berbere last updated on 12/May/24

$$ctg\left(x\right)=\frac{1}{y};y=tan\left(x\right)$$$$\iff Min(\frac{1}{8}(y+\frac{1}{y});y\in ]0,\mathrm{\infty }\left[\right)$$$$y+\frac{1}{y}⩾2\sqrt{y.\frac{1}{y}}=2$$$$min\left(\frac{1}{8}(y+\frac{1}{y})\right)⩾\frac{2}{8}=\frac{1}{4};$$$$equalityx=\frac{\pi }{4}$$

Commented by hardmath last updated on 12/May/24

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{dear}\mathrm{professor}\mathrm{cool}$$

Commented by Berbere last updated on 13/May/24

$$withepleasurimnotprofessor$$

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