a/ \(\dfrac{1}{x}+\dfrac{1}{y}\ge \dfrac{4}{x+y}\\\to \dfrac{x+y}{xy}\ge \dfrac{4}{x+y}\\\to \dfrac{(x+y)^2-4xy}{xy(x+y)}\\\to \dfrac{(x-y)^2}{xy(x+y)}\)
Vì \(x,y>0\)
\(\to \dfrac{(x-y)^2}{xy(x+y)}\ge 0\)
b/ Đặt \(A=\dfrac{1}{ac}+\dfrac{1}{bc}\)
Áp dụng BĐT Svac-xơ cho các số dương
\(\dfrac{1}{ac}+\dfrac{1}{bc}\ge \dfrac{4}{c(a+b)}\)
\(\to A\ge \dfrac{4}{c(2-c)}\\\to A\ge \dfrac{4}{-c^2+2c}\)
Ta có: \(-c^2+2c=-(c^2-2c+1)+1=-(c-1)^2+1\le 1\)
\(\to \dfrac{4}{-c^2+2c}\ge \dfrac{4}{1}=4\)
\(\to A\ge 4\)
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