Question Number 219587 by Nicholas666 last updated on 29/Apr/25
Commented by Nicholas666 last updated on 29/Apr/25
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Answered by Ghisom last updated on 29/Apr/25
$${a},\:{b}>\mathrm{0}\wedge{a}+{b}=\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow\:\mathrm{0}<{a},\:{b}<\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{let}\:{a}=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}+\mathrm{sin}\:\alpha\right)\wedge{b}=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{sin}\:\alpha\right) \\ $$$$\mathrm{inserting}\:\&\:\mathrm{transforming}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\frac{\mathrm{16}−\mathrm{6sin}^{\mathrm{2}} \:\alpha}{\mathrm{sin}^{\mathrm{4}} \:\alpha\:−\mathrm{20sin}^{\mathrm{2}} \:\alpha\:+\mathrm{64}}\leqslant\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha\:+\mathrm{2}}{\mathrm{8}} \\ $$$$\mathrm{with}\:\mathrm{sin}^{\mathrm{2}} \:\alpha\:=\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\alpha\:\mathrm{and}\:\mathrm{cos}\:\alpha\:={c}\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{{c}^{\mathrm{6}} +\mathrm{15}{c}^{\mathrm{4}} +\mathrm{39}{c}^{\mathrm{2}} −\mathrm{55}}{{c}^{\mathrm{4}} +\mathrm{18}{c}^{\mathrm{2}} +\mathrm{45}}\leqslant\mathrm{0} \\ $$$${c}^{\mathrm{4}} +\mathrm{18}{c}^{\mathrm{2}} +\mathrm{45}>\mathrm{0} \\ $$$${c}^{\mathrm{6}} +\mathrm{15}{c}^{\mathrm{4}} +\mathrm{39}{c}^{\mathrm{2}} −\mathrm{55}\leqslant\mathrm{0} \\ $$$$\left({c}^{\mathrm{2}} −\mathrm{1}\right)\left({c}^{\mathrm{2}} +\mathrm{5}\right)\left({c}^{\mathrm{2}} +\mathrm{11}\right)\leqslant\mathrm{0} \\ $$$$\left({c}^{\mathrm{2}} +\mathrm{5}\right)\left({c}^{\mathrm{2}} +\mathrm{11}\right)>\mathrm{0} \\ $$$${c}^{\mathrm{2}} −\mathrm{1}\leqslant\mathrm{0} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\alpha\:\leqslant\mathrm{1}\:\mathrm{true} \\ $$
Commented by Nicholas666 last updated on 29/Apr/25
$${thank}\:{you}\:{sir} \\ $$