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\(A\ge\frac{9}{2x+y+2y+z+2z+x}=\frac{9}{3\left(x+y+z\right)}=\frac{9}{3.3}=1\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Sửa đề:
\(\dfrac{x^2y}{x-1}+\dfrac{y^2z}{y-1}+\dfrac{z^2x}{z-1}=\dfrac{x^2y^2}{xy-y}+\dfrac{y^2z^2}{yz-z}+\dfrac{z^2x^2}{zx-x}\)
\(\ge\dfrac{\left(xy+yz+zx\right)^2}{xy+yz+zx-6}\)
Đặt \(t=xy+yz+zx>x+y+z=6\) thì ta có
\(\dfrac{t^2}{t-6}=24+\dfrac{t^2-24t+144}{t-6}=24+\dfrac{\left(t-12\right)^2}{t-6}\ge24\)
Vậy GTNN là 24 đạt dược khi \(x=y=z=2\)
Lời giải:Áp dụng BĐT Cauchy-Schwarz ta có:
$\frac{1}{2x+y+z}\leq \frac{1}{16}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$
$\frac{1}{x+2y+z}\leq \frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\right)$
$\frac{1}{x+y+2z}\leq \frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}\right)$
Cộng theo vế và rút gọn thì:
$\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\leq \frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$
\(VT=\dfrac{x^2}{x^2+2xy+3zx}+\dfrac{y^2}{y^2+2yz+3xy}+\dfrac{z^2}{z^2+2zx+3yz}\)
\(VT\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+5xy+5yz+5zx}=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+3\left(xy+yz+zx\right)}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\left(x+y+z\right)^2}=\dfrac{1}{2}\)
`P=x^3/(x+y)+y^3/(y+z)+z^3/(z+x)`
`=x^4/(x^2+xy)+y^4/(y^2+yz)+z^4/(z^2+zx)`
Ad bđt cosi-swart:
`P>=(x^2+y^2+z^2)^2/(x^2+y^2+z^2+xy+yz+zx)`
Mà `xy+yz+zx<=x^2+y^2+z^2)`
`=>P>=(x^2+y^2+z^2)^2/(2(x^2+y^2+z^2))=(x^2+y^2+z^2)/2=3/2`
Dấu "=" xảy ra khi `x=y=z=1`
`Q=(x^3+y^3)/(x+2y)+(y^3+z^3)/(y+2z)+(z^3+x^3)/(z+2x)`
`Q=(x^3/(x+2y)+y^3/(y+2z)+z^3/(z+2x))+(y^3/(x+2y)+z^3/(y+2z)+x^3/(z+2x))`
`Q=(x^4/(x^2+2xy)+y^4/(y^2+2yz)+z^4/(z^2+2zx))+(y^4/(xy+2y^2)+z^4/(yz+2z^4)+x^4/(xz+2x^2))`
Áp dụng BĐT cosi-swart ta có:
`Q>=(x^2+y^2+z^2)^2/(x^2+y^2+z^2+2xy+2yz+2zx)+(x^2+y^2+z^2)^2/(2(x^2+y^2+z^2)+xy+yz+zx))`
Mà`xy+yz+zx<=x^2+y^2+z^2`
`=>Q>=(x^2+y^2+z^2)^2/(3(x^2+y^2+z^2))+(x^2+y^2+z^2)^2/(3(x^2+y^2+z^2))=(2(x^2+y^2+z^2)^2)/(3(x^2+y^2+z^2))=(2(x^2+y^2+z^2))/3=2`
Dấu "=" xảy ra khi `x=y=z=1.`
2: Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=\dfrac{a\left(a+b+c\right)}{b+c}+\dfrac{b\left(a+b+c\right)}{c+a}+\dfrac{c\left(a+b+c\right)}{a+b}-a-b-c=\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c-a-b-c=0\)
1: Sửa đề: Cho \(x,y,z\ne0\) và \(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}=\dfrac{2}{2x+y+2z}\).
CM:....
Đặt 2x = x', 2z = z'.
Ta có: \(\dfrac{2}{x'}+\dfrac{2}{y}+\dfrac{2}{z'}=\dfrac{2}{x'+y+z'}\)
\(\Leftrightarrow\dfrac{1}{x'}+\dfrac{1}{y}+\dfrac{1}{z'}=\dfrac{1}{x'+y+z'}\)
\(\Leftrightarrow\dfrac{1}{x'}-\dfrac{1}{x'+y+z'}+\dfrac{1}{y}+\dfrac{1}{z'}=0\)
\(\Leftrightarrow\dfrac{y+z'}{x'\left(x'+y+z'\right)}+\dfrac{y+z'}{yz'}=0\)
\(\Leftrightarrow\dfrac{\left(y+z'\right)\left(yz'+x'^2+x'y+x'z'\right)}{x'yz'\left(x'+y+z'\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x'+y\right)\left(y+z'\right)\left(z'+x'\right)}{x'yz'\left(x'+y+z'\right)}=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(2z+2x\right)=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(z+x\right)=0\left(đpcm\right)\)
Áp dụng bđt phụ \(\dfrac{1}{A+B}\le\dfrac{1}{4}\left(\dfrac{1}{A}+\dfrac{1}{B}\right)\forall A,B>0\)
\(\dfrac{1}{2x+y+z}=\dfrac{1}{\left(x+y\right)+\left(x+z\right)}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}\right)\) Tương tự: \(\dfrac{1}{x+2y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(\Rightarrow\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=1\)
Dấu bằng xảy ra \(\Leftrightarrow x=y=z=\dfrac{3}{4}\)
Này Nguyễn Trọng Chiến, mk ko hiểu cái chỗ \(\dfrac{1}{4}.\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\le\dfrac{1}{16}.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}\right)\)??? Sao suy ra được vậy bn??
\(\dfrac{1}{x+x+y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{16}\left(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Tương tự: \(\dfrac{1}{x+2y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}\right)\) ; \(\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}\right)\)
Cộng vế với vế:
\(VT\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{3}{4}\)
Mk ko hiểu cái dòng đầu Nguyễn Việt Lâm Giáo viên, bn có thể nói chi tiết cách phân tích cho mk đc ko??
Ta có: \(\dfrac{16}{2x+y+z}\le\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)
\(\Leftrightarrow\dfrac{1}{2x+y+z}\le\dfrac{1}{16}\left(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\left(1\right)\)
Tương tự ta có: \(\left\{{}\begin{matrix}\dfrac{1}{x+2y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}\right)\left(2\right)\\\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}\right)\left(3\right)\end{matrix}\right.\)
Cộng (1), (2), (3) vế theo vế ta được:
\(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{4}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{4.4}{16}=1\)
Dấu = xảy ra khi \(x=y=z=\dfrac{3}{4}\)
\(B=\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\ge\frac{9}{2x+y+z+x+2y+z+x+y+2z}=\frac{9}{4\left(x+y+z\right)}\ge\frac{9}{4}.1=\frac{9}{4}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)