\(-\text{Theo bài ra: }D=\dfrac{x}{2x+y+z}+\dfrac{y}{2y+z+x}+\dfrac{z}{2z+x+y}\)

\(-\text{Đặt }\left\{{}\begin{matrix}a=2x+y+z\\b=2y+z+x\\c=2z+x+y\end{matrix}\right.\Rightarrow a+b+c=4\left(x+y+z\right)\)

\(\Rightarrow a-\dfrac{a+b+c}{4}=x\)

\(\Rightarrow x=\dfrac{3a-b-c}{4}\)

\(-\text{Tương tự: }\left\{{}\begin{matrix}y=\dfrac{3b-c-a}{4}\\z=\dfrac{3c-a-b}{4}\end{matrix}\right.\)

Suy ra \(D=\dfrac{3a-b-c}{4a}+\dfrac{3b-3c-a}{4b}+\dfrac{3c-a-b}{4c}\)

\(D=\dfrac{9}{4}-\left(\dfrac{b}{4a}+\dfrac{c}{4a}+\dfrac{c}{4b}+\dfrac{a}{4b}+\dfrac{a}{4c}+\dfrac{b}{4c}\right)\)

\(D=\dfrac{9}{4}-\dfrac{1}{4}\left[\left(\dfrac{b}{a}+\dfrac{a}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)+\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\right]\)

- Theo bất đẳng thức Cosi, ta có: \(\left\{{}\begin{matrix}\dfrac{b}{a}+\dfrac{a}{b}\ge2\\\dfrac{c}{a}+\dfrac{a}{b}\ge2\\\dfrac{c}{b}+\dfrac{b}{c}\ge2\end{matrix}\right.\)

Suy ra \( D\le\dfrac{9}{4}-\dfrac{1}{4}.6=\dfrac{9}{4}-\dfrac{6}{4}=\dfrac{3}{4}\)

Vậy \(D\le\dfrac{3}{4}\left(đpcm\right)\)

Cậu vào đây nha ! 

Câu hỏi của doanthihuong - Toán lớp 7 - Học toán với OnlineMath

Bài của lớp 7 ghê vậy!!

Áp dụng bất đẳng thức Cauchy cho 3 số dương x,y,z

ta có bổ đề \((a+b+c)({1\over a}+{1\over b}+{1\over c})\)  >  9

Áp dụng vào ta có

\(D*({2x+y+z\over x}+{2y+x+z\over y}+{2z+y+x\over z})\)  >9(1)

Ta có \({2x+y+z\over x}+{2y+x+z\over y}+{2z+y+x\over z}\) =\(2+{y+z\over x}+2+{z+x\over y}+2+{y+x\over z}\)=\(6-3+{y+z\over x}+1+{z+x\over y}+1+{y+x\over z}+1\)=\(3+{x+y+z\over x}+{y+x+z\over y}+{z+y+x\over z}\)=\(3+(x+y+z)({1\over x}+{1\over y}+{1\over z})\)  >  3+9=12

thay vào(1)

Ta có \(D \) <  \({9\over 12}\)=\({3\over 4}\) 

Dấu "=" xảy ra khi x=y=z 

=> ĐPCM

áp dụng bất đẳng thức phụ : \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

\(\frac{x}{2x+y+z}=\frac{x}{x+y+x+z}\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\)

\(\frac{y}{2y+x+z}\le\frac{1}{4}\left(\frac{y}{y+x}+\frac{y}{y+z}\right)\)

\(\frac{z}{2z+x+y}\le\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)

cộng vế theo vế

\(\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+x+y}\le\frac{1}{4}\left(\frac{x+y}{x+y}+\frac{y+z}{y+z}+\frac{z+x}{z+x}\right)=\frac{1}{4}\cdot3=\frac{3}{4}\)(đpcm)

\(\frac{x}{2x+y+z}=\frac{x}{\left(x+y\right)+\left(x+z\right)}\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\le\frac{1}{16}\left(\frac{x}{x}+\frac{x}{y}+\frac{x}{x}+\frac{x}{z}\right)=\frac{1}{16}\left(2+\frac{x}{y}+\frac{x}{z}\right)\)

\(tươngtự:\frac{y}{2y+z+x}\le\frac{1}{16}\left(2+\frac{y}{z}+\frac{y}{x}\right);\frac{z}{2z+x+y}\le\frac{1}{16}\left(2+\frac{z}{x}+\frac{z}{y}\right).\text{Cộng vế theo vế ta được:}\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+y+x}\le\frac{1}{16}\left(2+2+2+\frac{x}{y}+\frac{y}{x}+\frac{z}{x}+\frac{x}{z}+\frac{y}{z}+\frac{z}{y}\right)=\frac{1}{16}\left[6+\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{x}{z}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)\right]\ge\frac{1}{16}\left(6+2\sqrt{\frac{xy}{xy}}+2\sqrt{\frac{xz}{xz}}+2\sqrt{\frac{yz}{yz}}\right)=\)

\(=\frac{12}{16}=\frac{3}{4}\Rightarrow\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+x+y}\le\frac{3}{4}\left(\text{đpcm}\right)\)

Áp dụng tính chất : 1/a+b < = 1/4.(1/a+1/b) thì :

x/2x+y+z = x.(1/2x+y+z) = x.[1/(x+y)+(x+z)] < = x/4.(1/x+y + 1/x+z)

Tương tự : ..........

=> x/2x+y+z + y/x+2y+z + z/x+y+2z < = 1/4.(x/x+y + x/x+z + y/y+x + y/y+z + z/z+x + z/x+y )

                                                         = 1/4. [ ( x/x+y + y/x+y ) + ( y/y+z + z/z+y ) + ( z/z+x + x/x+z )

                                                         = 1/4.(1+1+1) = 3/4

Dấu "=" xảy ra <=> x=y=z

Vậy ..........

Tk mk nha

Đặt BT là P:

\(\text{P}=\frac{x}{\left(2x+y+z\right)}-1+\frac{y}{2y+z+x}-1+\frac{z}{\left(2z+x+y\right)}-1+3\)

\(\text{P}=-\frac{\left(x+y+z\right)}{\left(2x+y-z\right)}-\frac{\left(x+y+z\right)}{\left(2y+z+x\right)}-\frac{\left(x+y+z\right)}{\left(2z+x+y\right)}+3\)

\(\text{P}=-\left(x+y+z\right).\left[\frac{1}{\left(2x+y+z\right)}+\frac{1}{\left(2y+z+x\right)}+\frac{1}{\left(2z+x+y\right)}\right]+3\)

Co-si 3 số, ta có:

\(2x+y+z+2y+z+x+2z+x+y\ge3.\sqrt[3]{\left(2x+y+z\right)\left(2y+z+x\right)\left(2z+x+y\right)}\)

\(\Rightarrow4\left(x+y+z\right)\ge3.\sqrt[3]{\left(2x+y+z\right)\left(2y+z+x\right)\left(2z+x+y\right)}\)(1)

Co-si tiếp cho 3 số, ta có:

\(\frac{1}{\left(2x+y+z\right)}+\frac{1}{\left(2y+z+x\right)}+\frac{1}{\left(2z+x+y\right)}\ge3.\sqrt[3]{\frac{1}{\left(2x+y+z\right)}+\frac{1}{\left(2y+z+x\right)}+\frac{1}{\left(2z+x+y\right)}}\)(2)

Lấy (1) và (2) ta có: \(4\left(x+y+z\right)\left[\frac{1}{\left(2x+y+z\right)}+\frac{1}{\left(2y+z+x\right)}+\frac{1}{\left(2z+x+y\right)}\right]\ge9\)

\(\Rightarrow-\left(x+y+z\right).\left[\frac{1}{\left(2x+y+z\right)}+\frac{1}{\left(2y+z+x\right)}+\frac{1}{\left(2z+x+y\right)}\right]\le-\frac{9}{4}\)

Thay P, ta có:

\(\text{P}\le-\frac{9}{3}+3=\frac{3}{4}\left(ĐPCM\right)\)

Dấu "=" xảy ra khi x = y = z.

trả lời

x=y=z

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