Ta có:

\(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=6\ge\frac{9}{2\left(x+y+z\right)}\)\(\Rightarrow x+y+z\ge\frac{3}{4}\)

Lại có: \(\frac{1}{2x+3y+3z}=\frac{\left(\frac{3}{4}+\frac{1}{4}\right)^2}{2\left(x+y+z\right)+y+z}\le\frac{9}{32\left(x+y+z\right)}+\frac{1}{16\left(y+z\right)}\)

Do đó:

\(\frac{1}{2x+3y+3z}+\frac{1}{2y+3x+3z}+\frac{1}{2z+3x+3y}\)

\(\le\frac{9}{32\left(x+y+z\right)}\cdot3+\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)

\(\le\frac{9}{32\cdot\frac{3}{4}}+\frac{1}{16}\cdot6=\frac{3}{2}\)(Đpcm)

Tại sao \(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}=6\ge\frac{9}{2\left(x+y+z\right)}\)

Liên tục áp dụng bất đẳng thức \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\) và ta có:

\(\frac{1}{3x+3y+2x}=\frac{1}{2\left(x+y\right)+\left(x+y+2z\right)}\le\frac{1}{4}\left(\frac{1}{2\left(x+y\right)}+\frac{1}{\left(x+z\right)+\left(y+z\right)}\right)\le\frac{1}{8\left(x+y\right)}+\frac{1}{16}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)

Chứng minh tương tự tạ có:

\(\frac{1}{3x+2y+3z}\le\frac{1}{8\left(z+x\right)}+\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\)

\(\frac{1}{2x+3y+3z}\le\frac{1}{8\left(y+z\right)}+\frac{1}{16}\left(\frac{1}{z+x}+\frac{1}{x+y}\right)\)

Suy ra \(VT\le\frac{1}{8}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)+\frac{1}{8}\left(\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{z+x}\right)=\frac{3}{2}\)

Dấu "=" xảy ra <=> \(x=y=z=\frac{1}{4}\)

mơn ạ

Áp dụng \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(\frac{1}{3x+3y+2z}=\frac{1}{2\left(x+y\right)+\left(x+z\right)+\left(y+z\right)}\le\frac{1}{4}.\frac{1}{2\left(x+y\right)}+\frac{1}{4}.\frac{1}{x+z+y+z}\le\frac{1}{8\left(x+y\right)}+\frac{1}{4}.\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)

Áp dụng bất đẳng thức \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{\left(1+1+1+1\right)^2}{a+b+c+d}=\frac{16}{a+b+c+d}\)ta có :

\(\frac{16}{3x+3y+2z}\le\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\)

\(\frac{16}{3x+2y+3z}\le\frac{1}{x+z}+\frac{1}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\)

\(\frac{16}{2x+3y+3z}\le\frac{1}{y+z}+\frac{1}{y+z}+\frac{1}{x+y}+\frac{1}{x+z}\)

Cộng theo vế 3 đẳng thức trên ta được :

\(16.\left(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\right)\)

\(\le4.\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=4.6=24\)

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

Câu hỏi của NGUYỄN DOÃN ANH THÁI - Toán lớp 9 - Học toán với OnlineMath

a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)

Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2

b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)

Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)

Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)

đề đúng , giải sai kìa ...

trừ cho nhau là xong

Nói nghe có vẻ dễ ha Trần Hữu Ngọc Minh