Question Number 207361 by hardmath last updated on 12/May/24
$$\mathrm{y}\:=\:\frac{\mathrm{tg}\boldsymbol{\mathrm{x}}\:\:+\:\:\mathrm{ctg}\boldsymbol{\mathrm{x}}}{\mathrm{8}}\:\:\:\:\:,\:\:\:\:\:\left(\mathrm{0}\:;\:\frac{\pi}{\mathrm{2}}\right) \\ $$$$\mathrm{Find}:\:\:\:\mathrm{min}\left(\mathrm{y}\right)\:=\:? \\ $$
Answered by Berbere last updated on 12/May/24
![ctg(x)=(1/y);y=tan(x) ⇔Min((1/8)(y+(1/y));y∈]0,∞[) y+(1/y)≥2(√(y.(1/y)))=2 min((1/8)(y+(1/y)))≥(2/8)=(1/4); equality x=(π/4)](https://www.tinkutara.com/question/Q207367.png)
$${ctg}\left({x}\right)=\frac{\mathrm{1}}{{y}};{y}={tan}\left({x}\right) \\ $$$$\Leftrightarrow{Min}\left(\frac{\mathrm{1}}{\mathrm{8}}\left({y}+\frac{\mathrm{1}}{{y}}\right);{y}\in\right]\mathrm{0},\infty\left[\right) \\ $$$${y}+\frac{\mathrm{1}}{{y}}\geqslant\mathrm{2}\sqrt{{y}.\frac{\mathrm{1}}{{y}}}=\mathrm{2} \\ $$$${min}\left(\frac{\mathrm{1}}{\mathrm{8}}\left({y}+\frac{\mathrm{1}}{{y}}\right)\right)\geqslant\frac{\mathrm{2}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{4}}; \\ $$$${equality}\:{x}=\frac{\pi}{\mathrm{4}} \\ $$
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
$${withe}\:{pleasur}\:{im}\:{not}\:{professor} \\ $$