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NV
22 tháng 3 2021

\(VT=\dfrac{x^2}{x^2+2xy+3zx}+\dfrac{y^2}{y^2+2yz+3xy}+\dfrac{z^2}{z^2+2zx+3yz}\)

\(VT\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+5xy+5yz+5zx}=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+3\left(xy+yz+zx\right)}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\left(x+y+z\right)^2}=\dfrac{1}{2}\)

17 tháng 7 2021

 đặt\(A=\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\)

\(=>A=\dfrac{x^4}{2x^2+3xy+5xz}+\dfrac{y^4}{2y^2+3yz+5xy}+\dfrac{z^4}{2z^2+3xz+5yz}\)

BBDT AM-GM 

\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)

theo BDT AM -GM ta chứng minh được \(xy+yz+xz\le x^2+y^2+z^2\)

vì \(x^2+y^2\ge2xy\)

\(y^2+z^2\ge2yz\)

\(x^2+z^2\ge2xz\)

\(=>2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)< =>xy+yz+xz\le x^2+y^2+z^2\)

\(=>2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)\le10\left(x^2+y^2+z^2\right)\)

\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{10\left(x^2+y^2+z^2\right)}=\dfrac{x^2+y^2+z^2}{10}=\dfrac{\dfrac{1}{3}}{10}=\dfrac{1}{30}\left(đpcm\right)\)

dấu"=" xảy ra<=>x=y=z=1/3

18 tháng 12 2021

Áp dụng BĐT Cauchy-Schwarz:

\(\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\ge\dfrac{16}{3x+3y+2z}\\ \Leftrightarrow\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{2}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right)\\ \Leftrightarrow\sum\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x+y}+\dfrac{4}{y+z}+\dfrac{4}{z+x}\right)=\dfrac{4}{16}\cdot6=\dfrac{3}{2}\)

Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)

23 tháng 4 2017

Haha không giỡn nữa :v :focus:

Áp dụng BĐT Cauchy-Schwarz ta có:

\(L.H.S=Σ\dfrac{1}{2x+y+z}=7Σ\dfrac{1}{2\left(x+3y\right)+\left(y+3z\right)+4\left(z+3x\right)}\)

\(=\dfrac{1}{7}Σ\dfrac{\left(2+1+4\right)^2}{2\left(x+3y\right)+\left(y+3z\right)+4\left(z+3x\right)}\)

\(\le\dfrac{1}{7}Σ\left(\dfrac{2^2}{2\left(x+3y\right)}+\dfrac{1^2}{y+3z}+\dfrac{4^2}{4\left(z+3x\right)}\right)\)

\(=\dfrac{1}{7}Σ\left(\dfrac{2}{x+3y}+\dfrac{1}{y+3z}+\dfrac{4}{z+3x}\right)\)

\(=\dfrac{1}{7}Σ\dfrac{7}{x+3y}=Σ\dfrac{1}{x+3y}=R.H.S\)

23 tháng 4 2017

Áp dụng bất đẳng thức \(\dfrac{1}{x}+\dfrac{1}{y}\le\dfrac{4}{x+y}\) \(\forall x,y>0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x+3y}+\dfrac{1}{y+2z+x}\le\dfrac{4}{2x+4y+2z}=\dfrac{2}{x+2y+z}\\\dfrac{1}{y+3z}+\dfrac{1}{z+2x+y}\le\dfrac{4}{2x+2y+4z}=\dfrac{2}{x+y+2z}\\\dfrac{1}{z+3x}+\dfrac{1}{x+2y+z}\le\dfrac{4}{4x+2y+2z}=\dfrac{2}{2x+y+z}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{x+3y}+\dfrac{1}{y+3z}+\dfrac{1}{z+3x}+\dfrac{1}{y+2z+x}+\dfrac{1}{z+2x+y}+\dfrac{1}{x+2y+z}\le\dfrac{2}{x+2y+z}+\dfrac{2}{x+y+2z}+\dfrac{2}{2x+y+z}\)

\(\Rightarrow VT\le\left(\dfrac{2}{x+2y+z}-\dfrac{1}{x+2y+z}\right)+\left(\dfrac{2}{x+y+2z}-\dfrac{1}{y+x+2z}\right)+\left(\dfrac{2}{2x+y+z}-\dfrac{1}{z+2x+y}\right)\)

\(\Rightarrow VT\le\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}+\dfrac{1}{2x+y+z}\)

\(\Leftrightarrow\dfrac{1}{x+3y}+\dfrac{1}{y+3z}+\dfrac{1}{z+3x}\le\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}+\dfrac{1}{2x+y+z}\) ( đpcm )

5 tháng 12 2018

Sửa đề nhé\(\dfrac{1}{3x+3y+2z}=\dfrac{1}{\left(z+x\right)+\left(z+y\right)+\left(x+y\right)+\left(x+y\right)}\)

\(\le\dfrac{1}{16}\left(\dfrac{1}{x+z}+\dfrac{1}{z+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}\right)\)

CMTT và cộng theo vế:

\(VT\le\dfrac{1}{16}\left(\dfrac{1}{x+z}+\dfrac{1}{z+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{y+z}\right)\)

\(=\dfrac{1}{16}.24=\dfrac{3}{2}\)

\("="\Leftrightarrow x=y=z=\dfrac{1}{4}\)

24 tháng 11 2021

\(TH_1:x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\\ \Rightarrow Q=\dfrac{-z}{z}+\dfrac{-x}{x}+\dfrac{-y}{y}=-3\\ TH_2:x+y+z\ne0\\ \Rightarrow\dfrac{3x-2y+z}{x}=\dfrac{3y-2z+x}{y}=\dfrac{3z-2x+y}{z}=\dfrac{2x+2y+2z}{x+y+z}=2\\ \Rightarrow\left\{{}\begin{matrix}3x-2y+z=x\\3y-2z+x=y\\3z-2x+y=z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-2y=-z\\2y-2z=-x\\2z-2x=-y\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x-y=-\dfrac{z}{2}\\y-z=-\dfrac{x}{2}\\z-x=-\dfrac{y}{2}\end{matrix}\right.\)

\(\Rightarrow Q=-\dfrac{z}{2}:z-\dfrac{x}{2}:x-\dfrac{y}{2}:y=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}=-\dfrac{3}{2}\)

NV
31 tháng 12 2021

\(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}=\dfrac{x^4}{xy+2xz}+\dfrac{y^4}{yz+2xy}+\dfrac{z^4}{xz+2yz}\)

\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\) 

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)

10 tháng 11 2017

Ta có :

\(\dfrac{1}{3x+3y+2z}=\dfrac{1}{\left(2x+y+z\right)+\left(2y+x+z\right)}\)(1)

Áp dụng BĐT \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

\(\Rightarrow\left(1\right)\le\dfrac{1}{4}\left(\dfrac{1}{x+y+x+z}+\dfrac{1}{y+x+y+z}\right)\le\dfrac{1}{4}\left(\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{x+y}+\dfrac{1}{y+z}\right)\right)\)

\(=\dfrac{1}{16}\left(\dfrac{2}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)\)

tương tự với hai ông còn lại sau đó cộng lại ta được:

\(\Sigma\dfrac{1}{3x+3y+2z}\le\dfrac{24}{16}=\dfrac{3}{2}\)

28 tháng 2 2021

Áp dụng BĐT cosi ta có:

`x^6+y^6+z^6>=3root{3}{x^6y^6z^6}=3x^2y^2z^2`

`=>3x^2y^2z^2<=3`

`=>x^2y^2z^2<=1`

`=>xyz<=1`

`=>(x^3)/(yz)+(y^3)/(zx)+(z^3)/(xy)`

`=(x^4)/(xyz)+(y^4)/(xyz)+(z^4)/(xyz)>=x^4+y^4+z^4(@)`

Áp dụng BĐT bunhia với 2 cặp số `(x^2,y^2,z^2),(x,y,z)`

`=>(x^2+y^2+z^2)(x^4+y^4+z^4)>=(x^3+y^3+z^3)^3`

Mà `(x^3+y^3+z^3)^2>=3(x^3y^3+y^3z^3+z^3x^3)`

`=>(x^2+y^2+z^2)(x^4+y^4+z^4)>=3(x^3y^3+y^3z^3+z^3x^3)(@@)`

Áp dụng BĐT cosi ta có:

`x^6+1+1>=3root{3}{x^6}=3x^2`

`y^6+1+1>=3y^2`

`z^6+1+1>=3z^2`

`=>x^6+y^6+z^6+6>=3(x^2+y^2+z^2)`

`=>9>=3(x^2+y^2+z^2)`

`=>x^2+y^2+z^2<=3`

Kết hợp với `(@@)`

`=>(x^2+y^2+z^2)(x^4+y^4+z^4)>=(x^2+y^2+z^2)(x^3y^3+y^3z^3+z^3x^3)`

`=>x^4+y^4+z^4>=x^3y^3+y^3z^3+z^3x^3`

Kếp hợp với `(@)`

`=>(x^3)/(yz)+(y^3)/(zx)+(z^3)/(xy)>=x^3y^3+y^3z^3+z^3x^3`

Dấu = xảy ra khi `x=y=z=1`

28 tháng 2 2021

Học tốt nha ~.~