Câu 1:

\(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=x^2y^2+\frac{1}{256x^2y^2}+\frac{255}{256x^2y^2}+2\)

\(\ge\frac{1}{8}+2+\frac{255}{256x^2y^2}\)

Ta lại có: \(1=x+y\ge2\sqrt{xy}\Leftrightarrow1\ge16x^2y^2\)

\(\Rightarrow M\ge\frac{17}{8}+\frac{255}{16}=\frac{289}{16}\)

Dấu = xảy ra khi x=y=1/2

Áp dụng BDT Cauchy-Schwarz: \(\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)\ge\frac{1}{3x+3y+2z}\)

CMTT rồi cộng vế với vế ta có.\(VT\le\frac{1}{16}\cdot4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{3}{2}\)

Dấu = xảy ra khi x=y=z=1

Má mày giúp tao bài tao gửi đii:(

Ta có bất đẳng thức: với \(x,y>0\)

\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

Dấu \(=\)khi \(x=y\).

Áp dụng bất đẳng thức trên ta được: 

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

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

Tương tự với \(\frac{1}{3x+2y+3z},\frac{1}{3x+3y+2z}\)sau đó cộng lại vế với vế ta được: 

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

Dấu \(=\)xảy ra khi \(x=y=z=\frac{1}{8}\)

Đặt \(\left(y+z;x+z;x+y\right)=\left(a;b;c\right)\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{b+c-a}{2}\\y=\frac{a+c-b}{2}\\z=\frac{a+b-c}{2}\end{matrix}\right.\) và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2017}\)

\(P=\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\)

\(P\le\frac{1}{16}\left(\frac{2}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{2}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{2}{c}\right)\)

\(P\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{2017}{4}\)

Dấu "=" xảy ra khi \(a=b=c=6051\) hay \(x=y=z=\frac{6051}{2}\)

Áp dụng Bđt \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)

Ta có:

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

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

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

\(=\frac{1}{16}\left(6+\frac{1}{y+z}\right)\).Tương tự với 2 cái còn lại r` cộng lại ta đc:

\(P\le\frac{1}{16}\left[6+6+6+\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right]=\frac{3}{2}\)

Đặt \(\left\{{}\begin{matrix}x+y=a\\y+z=b\\x+z=c\end{matrix}\right.\) \(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)

\(P=\frac{1}{a+a+b+c}+\frac{1}{a+b+b+c}+\frac{1}{a+b+c+c}\)

\(\Rightarrow P\le\frac{1}{16}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)

\(\Rightarrow P\le\frac{1}{16}\left(\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\right)=\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{3}{2}\)

\(\Rightarrow P_{max}=\frac{3}{2}\) khi \(a=b=c=\frac{1}{2}\Rightarrow x=y=z=\frac{1}{4}\)

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

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