\(\Rightarrow\left(x+y+z\right)^2=1\Leftrightarrow x^2+y^2+z^2+2xy+2xz+2yz=1\)
\(P=\frac{3}{xy+xz+yz}+\frac{2}{x^2+y^2+z^2}=\frac{3\left(x+y+z\right)^2}{xy+xz+yz}+\frac{2\left(x^2+y^2+z^2\right)}{x^2+y^2+z^2}\)
\(P=\frac{3\left(x^2+y^2+z^2\right)+6\left(xy+xz+yz\right)}{xy+xz+yz}+\frac{2\left(x^2+y^2+z^2\right)+4\left(xy+xz+yz\right)}{x^2+y^2+z^2}\)
\(P=\frac{3\left(x^2+y^2+z^2\right)}{xy+xz+yz}+6+2+\frac{4\left(xy+xz+yz\right)}{x^2+y^2+z^2}\)
Áp dụng bất đẳng thức Cauchy ta có
\(\frac{3\left(x^2+y^2+z^2\right)}{xy+xz+yz}+\frac{4\left(xy+xz+yz\right)}{x^2+y^2+z^2}\ge2\sqrt{12}=4\sqrt{3}\)
\(\Rightarrow P\ge4\sqrt{3}+6+2=8+4\sqrt{3}\)
Dấu bằng thì bạn tự xét nhé
$\Rightarrow \left(x+y+z\right)^2=1\Leftrightarrow x^2+y^2+z^2+2xy+2xz+2yz=1$
$P=\frac{3}{xy+xz+yz}+\frac{2}{x^2+y^2+z^2}=\frac{3\left(x+y+z\right)^2}{xy+xz+yz}+\frac{2\left(x^2+y^2+z^2\right)}{x^2+y^2+z^2}$
$P=\frac{3\left(x^2+y^2+z^2\right)+6\left(xy+xz+yz\right)}{xy+xz+yz}+\frac{2\left(x^2+y^2+z^2\right)+4\left(xy+xz+yz\right)}{x^2+y^2+z^2}$
$P=\frac{3\left(x^2+y^2+z^2\right)}{xy+xz+yz}+6+2+\frac{4\left(xy+xz+yz\right)}{x^2+y^2+z^2}$
$\frac{3\left(x^2+y^2+z^2\right)}{xy+xz+yz}+\frac{4\left(xy+xz+yz\right)}{x^2+y^2+z^2}\ge2\sqrt{12}=4\sqrt{3}$
Sorry bạn!!!!!!!!!!!!!!!!!!
Mk mới hok lớp 8 thôi! =_=
Kb nhé! ^_^
Áp dụng BĐT Côsi:
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}.3.\sqrt[3]{\frac{1}{abc}}=9\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
Và \(xy+yz+zx\le\frac{1}{3}\left(x+y+z\right)^2=\frac{1}{3}\)
\(B=2\left(\frac{1}{x^2+y^2+z^2}+\frac{1}{xy+yz+zx}+\frac{1}{xy+yz+zx}\right)+\frac{1}{xy+yz+zx}\)
\(\ge\frac{2.9}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}+\frac{1}{\frac{1}{3}}=\frac{18}{\left(x+y+z\right)^2}+3=21\)
Dấu bằng xảy ra khi \(x=y=z=\frac{1}{3}\)
CHIU THUA!Lam sao ma giai duoc.HU WAA...
Mình tới không phải để trả lời mà là để nói khi nào có câu trả lời chính xác nhớ nói mình.
