Bất đẳng thức

Question

Cho

$x^2+y^2=1$ . Chứng minh rằng

$-6\leq \frac{2(x^2+6xy)}{x^2+2xy+3y^2}\leq 3$ .

score 7 · Answer 1

Ta có:

$2(x^2+6xy)+6(x^2+2xy+3y^2)=2(2x+3y)^2\ge0$

$\Rightarrow \frac{2(x^2+6xy)}{x^2+2xy+3y^2}\ge-6$
Dấu bằng xảy ra khi:

$\left\{ \begin{array}{l}2x+3y=0 \\ x^2+y^2=1 \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x=\frac{3}{\sqrt{13}}\\ y=\frac{-2}{\sqrt{13}} \end{array} \right.\\\left\{ \begin{array}{l} x=\frac{-3}{\sqrt{13}}\\ y=\frac{2}{\sqrt{13}} \end{array} \right. \end{array} \right.$

$2(x^2+6xy)-3(x^2+2xy+3y^2)=-(x-3y)^2\ge0$

$\Rightarrow \frac{2(x^2+6xy)}{x^2+2xy+3y^2} \le 3$
Dấu bằng xảy ra khi:

$\left\{ \begin{array}{l}x-3y=0 \\ x^2+y^2=1 \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x=\frac{3}{\sqrt{10}}\\ y=\frac{1}{\sqrt{10}} \end{array} \right.\\\left\{ \begin{array}{l} x=\frac{-3}{\sqrt{10}}\\ y=\frac{-1}{\sqrt{10}} \end{array} \right. \end{array} \right.$