Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
Ta có bđt \(\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)\)
\(\(\Rightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)\)
Áp dụng nhiều lần bđt trên ta được
\(\(\frac{1}{3x+3y+2z}=\frac{1}{\left(2x+y+z\right)+\left(x+2y+z\right)}\le\frac{1}{4}\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}\right)\)\)
\(\(\le\frac{1}{4}\left(\frac{1}{\left(x+y\right)+\left(x+z\right)}+\frac{1}{\left(x+y\right)+\left(y+z\right)}\right)\)\)
\(\(\le\frac{1}{4}\left[\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\right)\right]\)\)
\(\(\le\frac{1}{16}\left(\frac{2}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\right)\)\)
C/m tương tự cho các bđt còn lại
\(\(\frac{1}{3x+2y+3z}\le\frac{1}{16}\left(\frac{2}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\right)\)\)
\(\(\frac{1}{2x+3y+3z}\le\frac{1}{16}\left(\frac{2}{y+z}+\frac{1}{x+y}+\frac{1}{x+z}\right)\)\)
Cộng vế theo vế được
\(\(P\le\frac{1}{16}\left(\frac{4}{x+y}+\frac{4}{y+z}+\frac{4}{z+x}\right)=\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{1}{4}.6=\frac{3}{2}\)\)
Dấu "=" xảy ra
\(\(\Leftrightarrow\hept{\begin{cases}x=y=z\\\frac{1}{2x}+\frac{1}{2x}+\frac{1}{2x=6}\end{cases}}\)\)
\(\(\Leftrightarrow\hept{\begin{cases}x=y=z\\\frac{3}{2x}=6\end{cases}}\)\)
\(\(\Leftrightarrow\hept{\begin{cases}x=y=z\\x=\frac{1}{4}\end{cases}}\)\)
\(\(\Leftrightarrow x=y=z=\frac{1}{4}\)\)
Vậy ..........
cách khác :))
\(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\)\(\Leftrightarrow\)\(x+y+z\le3\)
\(P=\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\)
\(P=\frac{1}{3\left(x+y+z\right)-z}+\frac{1}{3\left(x+y+z\right)-y}+\frac{1}{3\left(x+y+z\right)-x}\)
\(\ge\frac{9}{9\left(x+y+z\right)-\left(x+y+z\right)}=\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.3}=\frac{3}{8}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\frac{1}{4}\)
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)
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 \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{3x+3y+2z}=\frac{1}{2\left(x+y\right)+\left(x+z\right)+\left(y+z\right)}\le\frac{1}{4}.\frac{1}{2\left(x+y\right)}+\frac{1}{4}.\frac{1}{x+z+y+z}\le\frac{1}{8\left(x+y\right)}+\frac{1}{4}.\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)
Áp dụng 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:
\(P\le\frac{1}{16}\left[6+6+6+\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right]=\frac{3}{2}\)
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 <=> \(x=y=z=\frac{1}{4}\)
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 \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)
Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{\left(1+1+1+1\right)^2}{a+b+c+d}=\frac{16}{a+b+c+d}\) ta có:
\(\frac{16}{2x+3y+3z}\le\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\)
\(\frac{16}{3x+2y+3z}\le\frac{1}{x+z}+\frac{1}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\)
\(\frac{16}{2x+3y+3z}\le\frac{1}{y+z}+\frac{1}{y+z}+\frac{1}{x+y}+\frac{1}{x+z}\)
Cộng theo vế 3 BĐT trên ta có:
\(16\left(\frac{1}{2x+3y+3z}+\frac{1}{3x+2y+3z}+\frac{1}{3x+3y+2z}\right)\)
\(\le4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)=4\cdot12=48\)
\(\Rightarrow\frac{1}{2x+3y+3z}+\frac{1}{3x+2y+3z}+\frac{1}{3x+3y+2z}\le3\)
các bạn có thể giúp mình giải bài toán này bằng bất đẳng thức \(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\le\frac{9}{2\left(x+y+z\right)}\)