Tìm tham số a để bất phương trình sau được thỏa mãn với mọi $x \in ( {0;\frac{\pi }{4}} )$
$sin^5x+cos^5x-a(sinx+cosx)\geq sinxcosx(sinx+cosx)$
Ta có
$c{\rm{o}}{{\rm{s}}^5}x + {\sin ^5}x = \left( {c{\rm{o}}{{\rm{s}}^2}{\rm{x}} + {\mathop{\rm s}\nolimits} {\rm{i}}{{\rm{n}}^2}{\rm{x}}} \right)\left( {c{\rm{o}}{{\rm{s}}^3}{\rm{x}} + {\mathop{\rm s}\nolimits} {\rm{i}}{{\rm{n}}^3}{\rm{x}}} \right) - \left( {{{\cos }^2}{\rm{x}}{\mathop{\rm s}\nolimits} {\rm{i}}{{\rm{n}}^3}{\rm{x}} + {{\cos }^3}{\rm{x}}{\mathop{\rm s}\nolimits} {\rm{i}}{{\rm{n}}^2}{\rm{x}}} \right)$
$ = \left( {c{\rm{osx +  sinx}}} \right)\left( {c{\rm{o}}{{\rm{s}}^2}{\rm{x}} + {\mathop{\rm s}\nolimits} {\rm{i}}{{\rm{n}}^2}{\rm{x}} - \cos {\rm{x}}{\mathop{\rm s}\nolimits} {\rm{inx}}} \right) - \frac{1}{4}{\sin ^2}2{\rm{x}}\left( {c{\rm{osx  +  sinx}}} \right)$
$ = \left( {\cos x + \sin x} \right)\left[ {1 - \frac{1}{2}{{\sin }^2}2{\rm{x}}} \right] - \frac{1}{4}{\sin ^2}2{\rm{x}}\left( {\cos x + \sin x} \right)$
$ = \left( {1 - \frac{{\sin 2{\rm{x}}}}{2} - \frac{{{{\sin }^2}2{\rm{x}}}}{4}} \right)\left( {c{\rm{osx}} + {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)$
Do đó bất phương trình đã cho trở thành:
$ \left( {1 - \frac{{\sin 2{\rm{x}}}}{2} - \frac{{{{\sin }^2}2{\rm{x}}}}{4}} \right)\left( {c{\rm{osx}} + {\mathop{\rm s}\nolimits} {\rm{inx}}} \right) - \frac{1}{2}\sin 2{\rm{x}}\left( {c{\rm{osx}} + {\mathop{\rm s}\nolimits} {\rm{inx}}} \right) \ge a\left( {c{\rm{osx}} + {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)$    $(1)$
Vì $x \in \left[ {0;\frac{\pi }{4}} \right]$ nên $\sin x + c{\rm{osx}} = \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) > 0$, do đó
$(1) \Leftrightarrow 1 - \sin 2{\rm{x  -  }}\frac{{{{\sin }^2}2{\rm{x}}}}{4} \ge a$
Do $1 - \sin 2{\rm{x  -  }}\frac{{{{\sin }^2}2{\rm{x}}}}{4} \ge 1 - 1 - \frac{1}{4} =  - \frac{1}{4}$ với mọi $x \in \left[ {0;\frac{\pi }{4}} \right]$
Dấu bằng đạt được khi $x = \frac{\pi }{4}$ nên ta có $ - \frac{1}{4} \ge a$

Đáp số : $a \le  - \frac{1}{4}$

Thẻ

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