Lời giải:

BĐT cần chứng minh tương đương với:

\((x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{9}{x+y+z}\right)\geq (x+y+z)\left(\frac{4}{x+y}+\frac{4}{y+z}+\frac{4}{z+x}\right)\)

\(\Leftrightarrow 12+\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}\geq 12+\frac{4x}{y+z}+\frac{4y}{x+z}+\frac{4z}{x+y}\)

\(\Leftrightarrow (\frac{y}{x}+\frac{y}{z}-\frac{4y}{x+z})+(\frac{z}{x}+\frac{z}{y}-\frac{4z}{x+y})+(\frac{x}{y}+\frac{x}{z}-\frac{4x}{y+z})\geq 0\)

\(\Leftrightarrow \frac{y(x-z)^2}{xz(x+z)}+\frac{z(x-y)^2}{xy(x+y)}+\frac{x(y-z)^2}{yz(y+z)}\geq 0\)

(luôn đúng với mọi $x,y,z>0$)

Do đó ta có đpcm.
Dấu "=" xảy ra khi $x=y=z$

Áp dụng công thức \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\left(x,y>0\right)\)

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

\(\frac{1}{y+z}\le\frac{1}{4y}+\frac{1}{4z}\)

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

Tương tự \(\hept{\begin{cases}\frac{1}{x+2y+z}\le\frac{1}{4}\left(\frac{1}{4x}+\frac{1}{2y}+\frac{1}{4z}\right)\left(2\right)\\\frac{1}{x+y+2z}\le\frac{1}{4}\left(\frac{1}{4x}+\frac{1}{4y}+\frac{1}{2z}\right)\left(3\right)\end{cases}}\)

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

=> \(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\)

Dấu "=" xảy ra <=> \(x=y=z=\frac{3}{4}\)

Áp dụng bất đẳng thức : \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)( với x , y > 0 )
Ta có : \(\frac{1}{2x+y+z}\le\frac{1}{4}\left(\frac{1}{2x}+\frac{1}{y+z}\right);\frac{1}{y+z}\le\frac{1}{4y}+\frac{1}{4z}\)

Suy ra : 

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

Tường tự ta có : 

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

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

Từ (1) , (2) và (3) 

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

\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\)

Dấu " = " xảy ra khi \(x=y=z=\frac{3}{4}\)

Chúc bạn học tốt !!!

địt mẹ laaaaaa

Sử dụng BĐT AM-GM, ta có: 

\(x^3+y^2\ge2yx\sqrt{x}\)

\(\Rightarrow\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2yx\sqrt{x}}=\frac{1}{xy}\)

Tương tự cộng lại suy ra: 

\(VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

Ghi chú: Này, mình mới lớp 6, nên giải chưa biết chắc là đúng hay sai nên lỡ có sai thì bạn đừng trách mình nhé!

Đặt \(A=\frac{x}{y\left(z+1\right)}+\frac{y}{z\left(x+1\right)}+\frac{z}{x\left(y+1\right)}\le\frac{9}{4}\)(Sửa đề)

Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)với a,b dương và x + y + z = 1,ta có:

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

Nhân hai vế với số dương xy, ta được:

\(\frac{4xy}{y\left(z+1\right)}\le\frac{4xy}{y}\left(\frac{1}{z+x}+\frac{1}{z+y}\right)\). Do đó:

\(4A=\frac{4xy}{y\left(z+1\right)}+\frac{4yz}{z\left(x+1\right)}+\frac{4zx}{x\left(y+1\right)}\)

\(\le\frac{4xy}{y}\left(\frac{1}{z+x}+\frac{1}{z+y}\right)+\frac{4yz}{z}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)+\frac{4zx}{x}\left(\frac{1}{y+z}+\frac{1}{y+z}\right)\)

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

\(=\frac{4x}{z+x}+\frac{4x}{z+y}+\frac{4y}{x+y}+\frac{4y}{x+z}+\frac{4z}{y+z}+\frac{4z}{y+z}\)

\(\Rightarrow4A\le\frac{4x+4y}{z+x}+\frac{4y+4z}{z+y}+\frac{4z+4x}{x+y}=x+y+z=9\)

Do : \(4A\le9\)nên \(A< \frac{9}{4}\)

Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Leftrightarrow\frac{xy+yz+zx}{xyz}=\frac{1}{x+y+z}\)

\(\Leftrightarrow\left(xy+yz+zx\right)\left(x+y+z\right)=xyz\)

\(\Leftrightarrow x^2y+xy^2+y^2z+yz^2+z^2x+zx^2+3xyz-xyz=0\)

\(\Leftrightarrow\left(x^2y+xy^2\right)+\left(yz^2+z^2x\right)+\left(zx^2+2xyz+y^2z\right)=0\)

\(\Leftrightarrow xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x+y\right)^2=0\)

\(\Leftrightarrow\left(x+y\right)\left(xy+z^2+yz+zx\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

=> x = -y hoặc y = -z hoặc z = -x

Không mất tổng quát giả sử x = -y, khi đó:

\(\frac{1}{x^{2015}}+\frac{1}{y^{2015}}+\frac{1}{z^{2015}}=-\frac{1}{y^{2015}}+\frac{1}{y^{2015}}+\frac{1}{z^{2015}}=\frac{1}{z^{2015}}\)

\(\frac{1}{x^{2015}+y^{2015}+z^{2015}}=\frac{1}{-y^{2015}+y^{2015}+z^{2015}}=\frac{1}{z^{2015}}\)

\(\Rightarrow\frac{1}{x^{2015}}+\frac{1}{y^{2015}}+\frac{1}{z^{2015}}=\frac{1}{x^{2015}+y^{2015}+z^{2015}}\)

\(x+y+z=0\)=>\(\frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}=0\)(*)

ta co :

\(\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}^2=\left|\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right|^2\)

\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}\)

\(\frac{2}{xy}+\frac{2}{xz}+\frac{2}{yz}=0\) luon dung vi (*)

=> dpcm

ban sua lai de di  dau "-"=>"+"

\(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\le\frac{1}{4x}+\frac{1}{4y}+\frac{1}{4z}+\frac{9}{4}\)

\(\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\le\left(x+y+z\right)\left(\frac{1}{4x}+\frac{1}{4y}+\frac{1}{4z}\right)+\frac{9}{4}\)

\(\Leftrightarrow\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x}\le\frac{y+z}{4x}+\frac{z+x}{4y}+\frac{x+y}{4z}\)

Ta có:

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

\(\ge\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=VT\)