Đê thiếu : Cho a,b,c > 0 ; a + b + c = 1 nhé
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Áp dụng BĐT Cô si cho 3 số dương
\(a+b+c\ge3\sqrt[3]{abc};\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\frac{1}{abc}}=9\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)
C2:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=9\)
Dấu "=" xảy ra tại \(a=b=c=\frac{1}{3}\)
C3:Nếu không muốn cm BĐT Cauchy-schwarz,ta dùng bđt phụ sau:\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Leftrightarrow\left(x-y\right)^2\ge0\left(true\right)\)
Áp dụng \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{4}{a+b}+\frac{1}{c}=\frac{4}{1-c}+\frac{1}{c}\)
Đến đây 1 biến thì ngon rồi
C4:\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3+\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)
\(=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)\)
\(\ge3+2+2+2=9\)
Dấu "=" xảy ra tại \(a=b=c=\frac{1}{3}\)
