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Question Number 126098 by ZiYangLee last updated on 17/Dec/20

Prove that 27(sin^4 α+2cos^4 α)≥16(sin2α)^2

$$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\mathrm{27}\left(\mathrm{sin}^{\mathrm{4}} \alpha+\mathrm{2cos}^{\mathrm{4}} \alpha\right)\geqslant\mathrm{16}\left(\mathrm{sin2}\alpha\right)^{\mathrm{2}} \\ $$

Answered by MJS_new last updated on 17/Dec/20

sin 2α =2sin α cos α 27(s^4 +2c^4 )≥64s^2 c^2 c=(√(1−s^2 )) 27(3s^4 −4s^2 +2)≥64s^2 (1−s^2 ) s^4 −((172)/(145))s^2 +((54)/(145))≥0 it′s easy to show this has no real zeros and for s=0 it′s true ⇒ always true

$$\mathrm{sin}\:\mathrm{2}\alpha\:=\mathrm{2sin}\:\alpha\:\mathrm{cos}\:\alpha \\ $$$$\mathrm{27}\left({s}^{\mathrm{4}} +\mathrm{2}{c}^{\mathrm{4}} \right)\geqslant\mathrm{64s}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} \\ $$$${c}=\sqrt{\mathrm{1}−{s}^{\mathrm{2}} } \\ $$$$\mathrm{27}\left(\mathrm{3}{s}^{\mathrm{4}} −\mathrm{4}{s}^{\mathrm{2}} +\mathrm{2}\right)\geqslant\mathrm{64}{s}^{\mathrm{2}} \left(\mathrm{1}−{s}^{\mathrm{2}} \right) \\ $$$${s}^{\mathrm{4}} −\frac{\mathrm{172}}{\mathrm{145}}{s}^{\mathrm{2}} +\frac{\mathrm{54}}{\mathrm{145}}\geqslant\mathrm{0} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{show}\:\mathrm{this}\:\mathrm{has}\:\mathrm{no}\:\mathrm{real}\:\mathrm{zeros} \\ $$$$\mathrm{and}\:\mathrm{for}\:{s}=\mathrm{0}\:\mathrm{it}'\mathrm{s}\:\mathrm{true}\:\Rightarrow\:\mathrm{always}\:\mathrm{true} \\ $$

Commented by talminator2856791 last updated on 17/Dec/20

very nice. quadratic is perfect.

$$\:\mathrm{very}\:\mathrm{nice}.\:\mathrm{quadratic}\:\mathrm{is}\:\mathrm{perfect}.\: \\ $$

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