Question

[LaYoLa]

cho x,y,z là các số dương . Tìm GTNN của biểu thức P=x/y+z   +    y/z+x     + z/x+y

Answer 1

[kudo shinichi]

\(P=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\)

\(\Rightarrow P+3=\frac{x}{y+z}+1+\frac{y}{z+x}+1+\frac{z}{x+y}+1\)

\(P+3=\frac{x+y+z}{y+z}+\frac{y+x+z}{z+x}+\frac{z+x+y}{x+y}\)

\(P+3=\left(x+y+z\right)\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)\)

\(\Rightarrow2P+6=\left[\left(x+y\right)+\left(y+z\right)+\left(z+x\right)\right]\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)\)

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~

Answer 2

[tth_new]

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:

\(P=\frac{x^2}{xy+xz}+\frac{y^2}{yz+xy}+\frac{z^2}{zx+yz}\ge\frac{\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\ge\frac{3\left(xy+yz+zx\right)}{2\left(xy+yz+zx\right)}=\frac{3}{2}^{\left(đpcm\right)}\)