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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