\(P=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\)
\(\Rightarrow P+3=\frac{x}{y+z}+1+\frac{y}{z+x}+1+\frac{z}{x+y}+1\)
\(P+3=\frac{x+y+z}{y+z}+\frac{y+x+z}{z+x}+\frac{z+x+y}{x+y}\)
\(P+3=\left(x+y+z\right)\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)\)
\(\Rightarrow2P+6=\left[\left(x+y\right)+\left(y+z\right)+\left(z+x\right)\right]\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)\)
Đặt \(\hept{\begin{cases}x+y=a\\y+z=b\\z+x=c\end{cases}}\)
\(\Rightarrow2P+6=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(2P+6=\left(\frac{a}{b}+\frac{b}{a}+1\right)+\left(\frac{b}{c}+\frac{c}{b}+1\right)+\left(\frac{c}{a}+\frac{a}{c}+1\right)\)
\(2P+6=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+3\)
\(\Rightarrow2P+3=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\)
Ta có: \(x;y;z>0\)
Áp dụng bất đẳng thức AM-GM ta có:
\(2P+3\ge2.\sqrt{\frac{a}{b}.\frac{b}{a}}+2.\sqrt{\frac{b}{c}.\frac{c}{b}}+2.\sqrt{\frac{c}{a}.\frac{a}{c}}=2+2+2=6\)
\(\Rightarrow2P\ge3\)
\(\Rightarrow P\ge1,5\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
Vậy \(P_{min}=1,5\Leftrightarrow a=b=c\)
Tham khảo nhé~
BĐT Nesbit à? =)))
ĐK: x,y,z > 0.Ta có: \(P=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\)
\(=\frac{x^2}{xy+xz}+\frac{y^2}{yz+xy}+\frac{z^2}{xz+yz}\). Áp dụng BĐT Cauchy - Schwarz,ta có:
\(P=\frac{x^2}{xy+xz}+\frac{y^2}{yz+xy}+\frac{z^2}{zx+yz}\ge\frac{\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\ge\frac{3\left(xy+yz+zx\right)}{2\left(xy+yz+zx\right)}=\frac{3}{2}^{\left(đpcm\right)}\)
Bạn bỏ cái chữ đpcm giúp mình nhé! Quen tay quá viết nhầm luôn :v
Và thêm câu cuối: "Vậy \(P_{min}=\frac{3}{2}\Leftrightarrow x=y=z\)"
Chết lại nhầm! Sau khi bỏ chữ đpcm bạn thêm vào dòng này:
"Vậy \(P_{min}=\frac{3}{2}\Leftrightarrow\frac{x}{y+z}=\frac{y}{z+x}=\frac{z}{x+y}\)
