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\(\frac{1}{3x+2y+z}=\frac{1}{x+x+x+y+y+z}\le\frac{1}{6^2}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\right)\)
\(=\frac{1}{36}\left(\frac{3}{x}+\frac{2}{y}+\frac{1}{z}\right)\)
Tương tự thì ta có:
\(\frac{1}{3x+2y+z}+\frac{1}{x+3y+2z}+\frac{1}{y+3z+2x}\)
\(\le\frac{1}{36}\left(\frac{3}{x}+\frac{2}{y}+\frac{1}{z}\right)+\frac{1}{36}\left(\frac{1}{x}+\frac{3}{y}+\frac{2}{z}\right)+\frac{1}{36}\left(\frac{1}{y}+\frac{3}{z}+\frac{2}{x}\right)\)
\(=\frac{6}{36}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{16}{6}=\frac{8}{3}\)
Dấu "=" xảy ra <=> x = y = z = 3/16
\(\frac{2+x}{x+1}+\frac{1-2y}{1+2y}\)
\(=1+\frac{1}{x+1}-1+\frac{2}{1+2y}\)
\(=\frac{1}{x+1}+\frac{1}{\frac{1}{2}+y}\)
Áp dụng BDDT AM-GM ta có:
\(\frac{1}{x+1}+\frac{1}{\frac{1}{2}+y}\ge\frac{2}{\sqrt{\left(x+1\right).\left(\frac{1}{2}+y\right)}}\ge\frac{4}{x+1+\frac{1}{2}+y}\ge\frac{4}{\frac{3}{2}+2}=\frac{4}{\frac{7}{2}}=\frac{8}{7}\)
Dấu " = " xảy ra <=> \(\frac{1}{x+1}=\frac{1}{\frac{1}{2}+y}\); x+y=2
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y=2\\y-x=\frac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\frac{5}{4}\\y=\frac{3}{4}\end{matrix}\right.\)