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Question-144936




Question Number 144936 by mathdanisur last updated on 30/Jun/21
Answered by mindispower last updated on 30/Jun/21
sec^4 (x)+csc^4 (x)≥(2/(cos^2 (x)sin^2 (x)))=(8/((sin(2x))^2 ))  ≥8  sec^6 (y)+csc^6 (y)≥(2/(cos^3 (y)sin^3 (y)))=((16)/(sin^3 (2y)))≥16  sec^8 (z)+csc^8 (z)≥(2/((cos(z)sin(z)^4 ))≥32  ⇔≥(8.16.32)^(1/3) =16  = if only if sin(2x)=sin(2y)=sin(2z)=1  ⇒x=y=z=(π/4)
$${sec}^{\mathrm{4}} \left({x}\right)+{csc}^{\mathrm{4}} \left({x}\right)\geqslant\frac{\mathrm{2}}{{cos}^{\mathrm{2}} \left({x}\right){sin}^{\mathrm{2}} \left({x}\right)}=\frac{\mathrm{8}}{\left({sin}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} } \\ $$$$\geqslant\mathrm{8} \\ $$$${sec}^{\mathrm{6}} \left({y}\right)+{csc}^{\mathrm{6}} \left({y}\right)\geqslant\frac{\mathrm{2}}{{cos}^{\mathrm{3}} \left({y}\right){sin}^{\mathrm{3}} \left({y}\right)}=\frac{\mathrm{16}}{{sin}^{\mathrm{3}} \left(\mathrm{2}{y}\right)}\geqslant\mathrm{16} \\ $$$${sec}^{\mathrm{8}} \left({z}\right)+{csc}^{\mathrm{8}} \left({z}\right)\geqslant\frac{\mathrm{2}}{\left({cos}\left({z}\right){sin}\left({z}\right)^{\mathrm{4}} \right.}\geqslant\mathrm{32} \\ $$$$\Leftrightarrow\geqslant\left(\mathrm{8}.\mathrm{16}.\mathrm{32}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{16} \\ $$$$=\:{if}\:{only}\:{if}\:{sin}\left(\mathrm{2}{x}\right)={sin}\left(\mathrm{2}{y}\right)={sin}\left(\mathrm{2}{z}\right)=\mathrm{1} \\ $$$$\Rightarrow{x}={y}={z}=\frac{\pi}{\mathrm{4}} \\ $$
Commented by mathdanisur last updated on 30/Jun/21
a lot cool Ser thanks
$${a}\:{lot}\:{cool}\:{Ser}\:{thanks} \\ $$

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