\(T=\dfrac{\left(xy\right)^2}{zx+zy}+\dfrac{\left(yz\right)^2}{xy+xz}+\dfrac{\left(zx\right)^2}{yx+yz}\ge\dfrac{xy+yz+zx}{2}\ge\dfrac{3}{2}\sqrt[3]{\left(xyz\right)^2}=\dfrac{3}{2}\)

\(B=\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\ge\frac{9}{2x+y+z+x+2y+z+x+y+2z}=\frac{9}{4\left(x+y+z\right)}\ge\frac{9}{4}.1=\frac{9}{4}\)

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

\(A\ge\frac{9}{2x+y+2y+z+2z+x}=\frac{9}{3\left(x+y+z\right)}=\frac{9}{3.3}=1\)

Dấu "=" xảy ra khi \(x=y=z=1\)

x+y+z=1

=>y+z=1-x; x+z=1-y; x+y=1-z

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

\(\frac{1+y}{1-y}=\frac{2y}{x+z}+1;\frac{1+z}{1-z}=\frac{2z}{x+y}+1\)

Do đó: \(\text{VT} = 2 \left( \frac{x}{y+z} + \frac{y}{x+z} + \frac{z}{x+y} \right) + 3\)

\(VP=\frac{2x}{y}+\frac{2y}{z}+\frac{2z}{x}\)

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

VT<=VP

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

=>\(\left(\frac{2x}{y}-\frac{2x}{y+z}\right)+\left(\frac{2y}{z}-\frac{2y}{x+z}\right)+\left(\frac{2z}{x}-\frac{2z}{x+y}\right)\ge3\)

=>\(2x \cdot \frac{z}{y(y+z)} + 2y \cdot \frac{x}{z(x+z)} + 2z \cdot \frac{y}{x(x+y)} \ge 3\)

=>\(\frac{xz}{y(y+z)}+\frac{xy}{z(x+z)}+\frac{yz}{x(x+y)}\ge\frac{3}{2}\quad(*)\)

Đặt \(T=\frac{xz}{y(y+z)}+\frac{xy}{z(x+z)}+\frac{yz}{x(x+y)}\)

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

Theo BĐT Cauchy, ta có:

\(T \ge \frac{(xy + yz + zx)^2}{xyz(y+z) + xyz(x+z) + xyz(x+y)} = \frac{(xy + yz + zx)^2}{2xyz(x+y+z)}\)

mà \(\left(xy+yz+xz\right)^2\ge3\left(xy+yz+xz\right)\)

nên \(T \ge \frac{3xyz(x+y+z)}{2xyz(x+y+z)} = \frac{3}{2}\) (ĐPCM)

Ta có bất đẳng thức AM-GM dạng phân thức sau: 

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Rightarrow \dfrac{1}{a+b}\le\dfrac{1}{4}(\dfrac{1}{a}+\dfrac{1}{b})\)

Dấu ''='' xảy ra khi và chỉ khi a=b

Quay lại bài toán: Áp dụng bđt trên, ta có:

\(\dfrac{1}{2x+y+z}=\dfrac{1}{(x+y)+(x+z)}\le\dfrac{1}{4}(\dfrac{1}{x+y}+\dfrac{1}{x+z})\\ \le\dfrac{1}{16}(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z})=\dfrac{1}{16}(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z})\)

Tương tự:

 \(\dfrac{1}{x+2y+z}\le\dfrac{1}{16}(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z})\); \(\dfrac{1}{x+y+2z}\le\dfrac{1}{16}(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z})\)

Cộng 3 phân thức lại, ta có:

\(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{1}{4}(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})=\dfrac{1}{4}.4=1\)

Dấu ''='' xảy ra khi và chỉ khi: \(x=y=z=\dfrac{3}{4}\)

Áp dụng BĐT BSC:

\(F=\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\)

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

\(=\dfrac{1}{16}\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{4}.4=1\)

\(maxF=1\Leftrightarrow x=y=z=\dfrac{3}{4}\)

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x+x+y+z)\geq (1+1+1+1)^2\)

\(\Rightarrow \frac{2}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{16}{2x+y+z}\)

Hoàn toàn tương tự:

\(\frac{1}{x}+\frac{2}{y}+\frac{1}{z}\geq \frac{16}{x+2y+z}\)

\(\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\geq \frac{16}{x+y+2z}\)

Cộng theo vế các BĐT vừa thu được:

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

\(\Rightarrow 16\geq 16\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)

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

Ta có đpcm.

Ta có :

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

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

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

Cộng từng vế của BĐT ta được :

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

Vậy BĐT đã được chứng minh !