Chủ đề này có 2 trả lời

[4869msnssk]

cho a,b,c dương thoả mãn $\sum \sqrt{a^2+b}^2=\sqrt{2k}$

chứng minh rằng  $\sum \frac{a^2}{b+c} \geq \frac{\sqrt{k}}{2}$

[Trang Luong]

cho a,b,c dương thoả mãn $\sum \sqrt{a^2+b}^2=\sqrt{2k}$

chứng minh rằng  $\sum \frac{a^2}{b+c} \geq \frac{\sqrt{k}}{2}$

Ta chứng minh BĐT phụ sau:

$P=2\left ( \frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b} \right )-\left [ \left ( \frac{a^2}{a+b}+\frac{b^2}{a+b} \right )+\left ( \frac{b^2}{b+c}+\frac{c^2}{b+c} \right )+\left ( \frac{c^2}{c+a}+\frac{a^2}{a+c} \right ) \right ]\geq 0$

 

Xét $\frac{2a^2}{b+c}-\frac{a^2}{a+c}-\frac{a^2}{a+b}=a^2\left ( \frac{2}{b+c}-\frac{1}{a+c}-\frac{1}{a+b} \right )=a^2.\frac{2a^2-b^2-c^2}{(a+b)(b+c)(a+c)}=\frac{2a^4-a^2b^2-a^2c^2}{(a+b)(b+c)(c+a)}$

 

$\Rightarrow P=\frac{2a^4-a^2b^2-a^2c^2+2b^4-b^2a^2-b^2c^2+2c^4-c^2a^2-c^2b^2}{(a+b)(b+c)(c+a)}\geq 0$

Theo BĐT Cauchy

Vậy $2\sum \frac{a^2}{b+c}\geq \frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{a+c}+\frac{b^2+a^2}{a+b}\geq \sum \frac{b^2+c^2}{\sqrt{2\left ( b^2+c^2 \right )}}=\frac{1}{\sqrt{2}}\left ( \sum \sqrt{a^2+b^2} \right )\Rightarrow \sum \frac{a^2}{b+c}\geq \frac{\sqrt{k}}{2}$

Bài viết đã được chỉnh sửa nội dung bởi khonggiadinh: 03-03-2014 - 21:23

[Vu Thuy Linh]

Ta chứng minh BĐT phụ sau:

$P=2\left ( \frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b} \right )-\left [ \left ( \frac{a^2}{a+b}+\frac{b^2}{a+b} \right )+\left ( \frac{b^2}{b+c}+\frac{c^2}{b+c} \right )+\left ( \frac{c^2}{c+a}+\frac{a^2}{a+c} \right ) \right ]\geq 0$

 

Xét $\frac{2a^2}{b+c}-\frac{a^2}{a+c}-\frac{a^2}{a+b}=a^2\left ( \frac{2}{b+c}-\frac{1}{a+c}-\frac{1}{a+b} \right )=a^2.\frac{2a^2-b^2-c^2}{(a+b)(b+c)(a+c)}=\frac{2a^4-a^2b^2-a^2c^2}{(a+b)(b+c)(c+a)}$

 

$\Rightarrow P=\frac{2a^4-a^2b^2-a^2c^2+2b^4-b^2a^2-b^2c^2+2c^4-c^2a^2-c^2b^2}{(a+b)(b+c)(c+a)}\geq 0$

Theo BĐT Cauchy

Vậy $2\sum \frac{a^2}{b+c}\geq \frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{a+c}+\frac{b^2+a^2}{a+b}\geq \sum \frac{b^2+c^2}{\sqrt{2\left ( b^2+c^2 \right )}}=\frac{1}{\sqrt{2}}\left ( \sum \sqrt{a^2+b^2} \right )\Rightarrow \sum \frac{a^2}{b+c}\geq \frac{\sqrt{k}}{2}$

Cách khác:

$P=\sum \frac{a^{2}}{b+c}\geq \sum \frac{a^{2}}{\sqrt{2(b^{2}+c^{2})}}$

Đặt $\sqrt{a^{2}+b^{2}}=x, {b^{2}+c^{2}}=y,{c^{2}+a^{2}}=z$ => $x+y+z=\sqrt{2k}$

Ta có: $\sum \frac{a^{2}}{b+c}\geq \sum \frac{x^{2}+z^{2}-y^{2}}{2\sqrt{2y^{2}}}=\sum \frac{1}{2\sqrt{2}}.(\sum \frac{x^{2}+z^{2}-y^{2}}{y})\geq \frac{1}{2\sqrt{2}}.(\sum \frac{(x+z)^{2}}{2y}-y)$

Mà $\sum \left (\frac{(x+z)^{2}}{2y}-y \right )=\sum \left (\frac{(x+z)^{2}}{2y}+2y-3y \right )\geq \sum \left (2(x+z)-3y \right )$ = $x+y+z=\sqrt{2k}$

Suy ra P $\geq \frac{\sqrt{k}}{2}$