Bài 103082

Question

$a/$Cho $\begin{cases}x>0 \\ t>1 \end{cases}.$Chứng minh rằng: $x^{t}-tx+t-1\geq 0$
$b/$Chứng minh rằng trong $\triangle ABC:$
$\left ( \tan \frac{A}{2} \right )^{2\sqrt{2}}+\left ( \tan \frac{B}{2} \right )^{2\sqrt{2}}+\left ( \tan \frac{C}{2} \right )^{2\sqrt{2}}\geq 3^{1-\sqrt{2}}$

score 0 · Answer 1

$a/$Do $x-1>-1, t>1$,theo BĐT Bernoulli:
$x^{t}=[1+\left ( x-1 \right )]^{t}\geq 1+t\left ( x-1 \right )=1+tx-t $
$\Rightarrow x^{t}-tx+t-1\geq 0$
Dấu "=" xảy ra $\Leftrightarrow x=1$
$b/$Chọn $\begin{cases}x= 3\tan^{2} \frac{A}{2}>0 \\ t= \sqrt{2}\end{cases}$
Theo câu $a $ $\Rightarrow \left ( 3\tan^{2} \frac{A}{2} \right )^{ \sqrt{2}}-3\sqrt{2}\tan^{2} \frac{A}{2}+\sqrt{2}-1\geq 0$
$\Leftrightarrow \left ( 3\tan^{2} \frac{A}{2} \right )^{ \sqrt{2}}\geq 3\sqrt{2}\tan^{2} \frac{A}{2}-\sqrt{2}+1$ $\left ( 1 \right )$
Tương tự ta cũng có:
$ \left ( 3\tan^{2} \frac{B}{2} \right )^{ \sqrt{2}}\geq 3\sqrt{2}\tan^{2} \frac{B}{2}-\sqrt{2}+1$ $\left ( 2 \right )$
$\left ( 3\tan^{2} \frac{C}{2} \right )^{ \sqrt{2}}\geq 3\sqrt{2}\tan^{2} \frac{C}{2}-\sqrt{2}+1$ $\left ( 3 \right )$
Cộng $\left ( 1 \right )$,$\left ( 2 \right )$,$\left ( 3 \right )$ vế với vế:
$\Rightarrow 3^\sqrt{2}[ \left ( \tan \frac{A}{2} \right )^{2\sqrt{2}}+\left ( \tan \frac{B}{2} \right )^{2\sqrt{2}}+\left ( \tan \frac{C}{2} \right )^{2\sqrt{2}}]\geq$
$\geq 3\sqrt{2}\left ( \tan^{2} \frac{A}{2}+\tan^{2} \frac{B}{2}+\tan^{2} \frac{C}{2} \right )-3\sqrt{2}+3$
$\geq 3\sqrt{2}.1-3\sqrt{2}+3=3$
(vì $ \tan^{2} \frac{A}{2}+\tan^{2} \frac{B}{2}+\tan^{2} \frac{C}{2}\geq \tan \frac{A}{2}.\tan \frac{B}{2}+\tan \frac{B}{2}.\tan \frac{C}{2}+\tan \frac{C}{2}.\tan \frac{A}{2}=1$ )
$\Rightarrow \left ( \tan \frac{A}{2} \right )^{2\sqrt{2}}+\left ( \tan \frac{B}{2} \right )^{2\sqrt{2}}+\left ( \tan \frac{C}{2} \right )^{2\sqrt{2}}\geq 3^{1-\sqrt{2}}$
Dấu "=" xảy ra $\Leftrightarrow 3\tan^{2} \frac{A}{2}=3\tan^{2} \frac{B}{2}=3\tan^{2} \frac{C}{2}=1$
$\Leftrightarrow \tan \frac{A}{2}=\tan \frac{B}{2}=\tan \frac{C}{2}=\frac{1}{ \sqrt{3}}\Leftrightarrow A=B=C=\frac{\Pi}{3}$
$\Leftrightarrow \triangle ABC$ đều

Bài 103082

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Bài 103082

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