Đề bài: $a/$Chứng minh rằng:$\left ( x+ y\right )^{2}-xy+1\geq \left ( x +y\right )\sqrt{3},\forall x,y$$b/$Cho $\triangle ABC$.Chứng minh rằng: $\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}\geq \sqrt{3}$

Lời giải

Đề bài:
$a/$Chứng minh rằng:$\left ( x+ y\right )^{2}-xy+1\geq \left ( x +y\right )\sqrt{3},\forall x,y$$b/$Cho $\triangle ABC$.Chứng minh rằng: $\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}\geq \sqrt{3}$
Lời giải

$a/$$\left ( x+ y\right )^{2}-xy+1\geq \left ( x +y\right )\sqrt{3}$
$\Leftrightarrow x^{2}+\left ( y- \sqrt{3}\right )x+y^{2}-\sqrt{3}y+1\geq 0$
$\Leftrightarrow x^{2}+\left ( y- \sqrt{3}\right )x+\frac{\left ( y- \sqrt{3}\right )^{2}}{4}+y^{2}-\sqrt{3}y+1-\frac{\left ( y- \sqrt{3}\right )^{2}}{4}\geq 0$
$\Leftrightarrow x^{2}+\left ( y- \sqrt{3}\right )x+\frac{\left ( y- \sqrt{3}\right )^{2}}{4}+\frac{3y^{2}-2\sqrt{3}y+1}{4}\geq 0$
$\Leftrightarrow \left ( x +\frac{ y- \sqrt{3}}{2}\right )^{2}+\frac{\left ( \sqrt{3}y-1 \right )^{2}}{4}\geq 0$
Luôn đúng $\Rightarrow $ (ĐPCM)
$b/$Đặt:$ \begin{cases}x=\tan \frac{A}{2} >0\\ y= \tan \frac{B}{2} >0\end{cases}$
$\tan \frac{C}{2} =\cot \left ( \frac{\Pi}{2}-\frac{C}{2}\right )=\cot \left ( \frac{A}{2}+\frac{B}{2}\right )=\frac{1}{ \tan \left ( \frac{A}{2}+\frac{B}{2}\right )}=\frac{1-xy}{x+y}$
Theo câu a ta có:
$\left ( x+ y\right )^{2}+1-xy\geq \left ( x +y\right )\sqrt{3}$
$\Rightarrow \left ( x+ y\right )^{2}+\left (\tan \frac{C}{2}  \right )\left ( x +y\right )\geq \left ( x +y\right )\sqrt{3}$
$\Rightarrow x+y+\tan \frac{C}{2}\geq \sqrt{3}$
$\Rightarrow \tan \frac{A}{2} +\tan \frac{B}{2}+\tan \frac{C}{2}\geq \sqrt{3}$ $\Rightarrow $(ĐPCM)

=========
Chuyên mục: Các dạng bất đẳng thức khác