Trích:
Nguyên văn bởi khucyeuthuong  Bài 25: Cho $f\left( \tan 2x \right)={{\tan }^{4}}x+\frac{1}{{{\tan }^{4}}x} với x\ \in \left( 0\ ;\ \frac{\pi }{4} \right)$. Tìm giá trị lớn nhất $f(\sin x) + f(\cos x)$ |
Đặt $t = \tan 2x$ suy ra $ t=\frac{2\tan x}{1-{{\tan }^{2}}x}$ nên $\frac{2}{t}=\frac{1}{\tan x}-\tan x$ $\Rightarrow \frac{4}{{{t}^{2}}}=\frac{1}{{{\tan }^{2}}x}+{{\tan }^{2}}x-2$
$\Rightarrow {{\left( \frac{4}{{{t}^{2}}}+2 \right)}^{2}}=\frac{1}{{{\tan }^{4}}x}+{{\tan }^{4}}x-4$ hay ${{\tan }^{4}}x+\frac{1}{{{\tan }^{4}}x}=\frac{16}{{{t}^{4}}}+\frac{16}{{{t}^{2}}} +2 = f(t)$
Nên: $f\left( \sin x \right)+f\left( \cos x \right)=16\left( \frac{1}{{{\sin }^{4}}x}+\frac{1}{{{\cos }^{4}}x} \right)+16\left( \frac{1}{{{\sin }^{2}}x}+\frac{1}{{{\cos }^{2}}x} \right)+4$
$f\left( \sin x \right)+f\left( \cos x \right)=16.\frac{8}{{{\sin }^{2}}2x}+16.\frac{4}{\sin 2x}+4 \geq 196$
[RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT]