Cho x+y+z=0 ; x+1>0 ; y+1>0 ; z+4>0 . Tìm GTNN x/x+1 + y/y+1 + z/z+4
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19 tháng 1 2022
Đặt \(\dfrac{x-y}{z}=m,\dfrac{y-z}{x}=n,\dfrac{z-x}{y}=p\), ta có:
\(\left(m+n+p\right)\left(\dfrac{1}{m}+\dfrac{1}{n}+\dfrac{1}{p}\right)=3+\dfrac{n+p}{m}+\dfrac{p+m}{n}+\dfrac{m+n}{p}\)
Tính \(\dfrac{n+p}{m}\) theo x, y, z ta được:
\(\dfrac{n+p}{m}=\dfrac{z}{x-y}.\dfrac{y^2-yz+xz-x^2}{xy}=\dfrac{z}{xy}\left(-x-y+x\right)\)
\(=\dfrac{z}{xy}\left(-x-y-z+2z\right)=\dfrac{2x^2}{xy}\) vì \(\left(x+y+z\right)=0\)
Tương tự: \(\dfrac{m+p}{n}=\dfrac{2x^2}{yz}.\dfrac{m+n}{p}=\dfrac{2y^2}{xz}\)
Vậy \(\left(m+n+p\right)\left(\dfrac{1}{m}+\dfrac{1}{n}+\dfrac{1}{p}\right)=3+\dfrac{2\left(x^3+y^3+z^3\right)}{xyz}=3+\dfrac{2.3xyz}{xyz}=3+6=9\)
23 tháng 10 2021
Đặt \(P=\left(\dfrac{x-y}{z}+\dfrac{y-z}{x}+\dfrac{z-x}{y}\right)\left(\dfrac{z}{x-y}+\dfrac{x}{y-z}+\dfrac{y}{z-x}\right)=9\)
Đặt \(\left\{{}\begin{matrix}\dfrac{x-y}{z}=a\\\dfrac{y-z}{x}=b\\\dfrac{x-z}{y}=c\end{matrix}\right.\)
\(\Leftrightarrow P=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\\ =1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\\ =3+\dfrac{a+c}{b}+\dfrac{a+b}{c}+\dfrac{b+c}{a}\)
Ta có \(\dfrac{a+c}{b}=\dfrac{\dfrac{x-y}{z}+\dfrac{z-x}{y}}{\dfrac{y-z}{x}}=\dfrac{xy-y^2+z^2-xz}{yz}\cdot\dfrac{x}{y-z}\)
\(=\dfrac{\left(z-y\right)\left(y+z-x\right)x}{yz\left(y-z\right)}=\dfrac{x\left(x-y-z\right)}{yz}\)
Mà \(x+y+z=0\Leftrightarrow x=-y-z\)
\(\Leftrightarrow\dfrac{a+c}{b}=\dfrac{x\left(x+x\right)}{yz}=\dfrac{2x^2}{yz}\)
Cmtt ta được \(\dfrac{a+b}{c}=\dfrac{2y^2}{xz};\dfrac{b+c}{a}=\dfrac{2z^2}{xy}\)
Cộng vế theo vế
\(\Leftrightarrow P=\dfrac{2x^2}{yz}+\dfrac{2y^2}{xz}+\dfrac{2z^2}{xy}+3=\dfrac{2x^3+2y^3+2z^3}{xyz}+3\\ \Leftrightarrow P=\dfrac{2\left(x^3+y^3+z^3\right)}{xyz}+3\)
Lại có \(x+y+z=0\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\\ \Leftrightarrow x^3+y^3+z^3=3xyz\)
Thế vào \(P\)
\(\Leftrightarrow P=\dfrac{2\cdot3xyz}{xyz}+3=6+3=9\)
TL
0
18 tháng 3 2020
Ta có: \(x+y+z=0\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\left(y+z\right)\\y=-\left(z+x\right)\\z=-\left(x+y\right)\end{matrix}\right.\)
Đặt \(A=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}+3\)
Thay \(x=-\left(y+z\right),\) \(y=-\left(z+x\right),\) \(z=-\left(x+y\right)\) vào A, ta có:
\(A=\frac{y+z}{-\left(y+z\right)}+\frac{z+x}{-\left(z+x\right)}+\frac{x+y}{-\left(x+y\right)}+3\)
\(\Leftrightarrow A=\left(-1\right)+\left(-1\right)+\left(-1\right)+3\)
\(\Leftrightarrow A=-3+3\)
\(\Leftrightarrow A=0\) ( ĐPCM )
15 tháng 2 2020
Áp dụng bđt AM-GM ta có:
\(\frac{x^2}{x+y}+\frac{x+y}{4}\ge2\sqrt{\frac{x^2}{x+y}.\frac{x+y}{4}}=x\)
\(\frac{y^2}{x+z}+\frac{x+z}{4}\ge2\sqrt{\frac{y^2}{x+z}.\frac{x+z}{4}}\ge y\)
\(\frac{z^2}{x+y}+\frac{x+y}{4}\ge2\sqrt{\frac{z^2}{x+y}.\frac{x+y}{4}}\ge z\)
Cộng từng vế các bđt trên ta được:
\(P+\frac{x+y+z}{2}\ge x+y+z\)
\(\Rightarrow P\ge\frac{x+y+z}{2}=1\)
Dấu"="xảy ra \(\Leftrightarrow x=y=z=1\)
Vậy Min P=1 \(\Leftrightarrow x=y=z=1\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 7 2021
\(x^2+y^2-z^2>0\Rightarrow x^2+2xy+y^2-z^2>0\)
\(\Rightarrow\left(x+y\right)^2-z^2>0\)
\(\Rightarrow\left(x+y-z\right)\left(x+y+z\right)>0\)
Mà x;y;z>0 \(\Rightarrow x+y+z>0\)
\(\Rightarrow x+y-z>0\)
13 tháng 12 2021
\(P=\sum\dfrac{1}{x+y+1}\ge\dfrac{9}{2\left(x+y+z\right)+3}=\dfrac{9}{2.1+3}=\dfrac{9}{5}\)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)
TN
1
21 tháng 8 2015
Xét hiệu: (x+y)(y+z)(z+x)-8xyz=0
(=) (x+y)>=2√xy
(y+z)>=2√yz
(z+x)>=2√zx
(=) (x+y)(y+z)(z+x)>=8√x^2 y^2 z^2
(=) (x+y)(y+z)(x+z)>=8|x| |y| |z|
(=) ( x+y)(y+z)(z+x)>= 8xyz
Ta có:
\(\frac{x}{x+1}=1-\frac{1}{x+1}\)
\(\frac{y}{y+1}=1-\frac{y}{y+1}\)
\(\frac{z}{z+4}=1-\frac{4}{z+4}\)
\(\Rightarrow\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+4}=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{4}{z+4}\right)\)
\(\le\left[3-\left(\frac{4}{x+y+2}+\frac{4}{z+4}\right)\right]\le\left(3-\frac{16}{x+y+z+6}\right)=3-\frac{16}{6}=\frac{1}{3}\)