Đề bài: Cho  $\begin{cases}s,t,u,v \in (0;\frac{\pi}{2}) \\ s+t+u+v=\pi \end{cases}$Chứng minh rằng:  $\frac{\sqrt{2}\sin s-1}{\cos s}+\frac{\sqrt{2}\sin t-1}{\cos t}+\frac{\sqrt{2}\sin u-1}{\cos u}+\frac{\sqrt{2}\sin v-1}{\cos v}\geq 0$

Lời giải

Đề bài:
Cho  $\begin{cases}s,t,u,v \in (0;\frac{\pi}{2}) \\ s+t+u+v=\pi \end{cases}$Chứng minh rằng:  $\frac{\sqrt{2}\sin s-1}{\cos s}+\frac{\sqrt{2}\sin t-1}{\cos t}+\frac{\sqrt{2}\sin u-1}{\cos u}+\frac{\sqrt{2}\sin v-1}{\cos v}\geq 0$
Lời giải

Đặt :  $a=\tan s,   b=\tan t,    c=\tan u,      d=\tan v   (a,b,c,d>0)$
Khi đó, từ    $s+t=\pi-(u+v)   \Rightarrow  \tan (s+t)+\tan (u+v)=0$
$\Leftrightarrow  \frac{a+b}{1-ab}+\frac{c+d}{1-cd}=0$
$\Leftrightarrow (a+b)(1-cd)+(c+d)(1-ab)=0$
$\Leftrightarrow  a+b+c+d=abc+abd+cda+cdb$
$\Leftrightarrow (a+b)(a+c)(a+d)=(a^2+a(b+c)+bc)(a+d)$
$= a^3+a^2(b+c)+abc+a^2d+ad(b+c)+bcd$
$=a^2(a+b+c+d)+abc+abd+acd+bcd$
$=(a^2+1)(a+b+c+d)$
$\Rightarrow  \frac{a^2+1}{a+b}=\frac{(a+c)(a+d)}{a+b+c+d}$
Theo BĐT Bunhiacopski:
$(a+b+c+d)^2=[(a+b)+(b+c)+(c+d)+(d+a)].$
                                                                 $ .\left ( \frac{a^2+1}{a+b}+ \frac{b^2+1}{b+c}+ \frac{c^2+1}{c+d}+ \frac{d^2+1}{d+a} \right )$
           $\geq  (\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}+\sqrt{d^2+1})^2$
$\Rightarrow   \sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}+\sqrt{d^2+1} \leq \sqrt{2}(a+b+c+d)$
$\Rightarrow  \frac{1}{\cos s}+ \frac{1}{\cos t}+ \frac{1}{\cos u}+ \frac{1}{\cos v} \leq \sqrt{2}(\tan s+\tan t+\tan u+\tan v)$
$\Rightarrow$  đpcm

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Chuyên mục: Bất đẳng thức Bunhiacốpxki