Question

cho a,b,c>o ,a+b+c=1

cmr \(\dfrac{1}{ab}+\dfrac{1}{bc}\ge16\)

Answer 1

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{1}{ab}+\frac{1}{bc}\right)(ab+bc)\geq 4\)

\(\Rightarrow \frac{1}{ab}+\frac{1}{bc}\geq \frac{4}{ab+bc}=\frac{4}{b(a+c)}(1)\)

Áp dụng BĐT AM-GM:

\(b(a+c)\leq \left(\frac{b+a+c}{2}\right)^2=\frac{1}{4}(2)\)

Từ \((1);(2)\Rightarrow \frac{1}{ab}+\frac{1}{bc}\geq \frac{4}{b(a+c)}\geq \frac{4}{\frac{1}{4}}=16\) (đpcm)

Dấu "=" xảy ra khi \(b=a+c; \frac{1}{ab}=\frac{1}{bc}\Rightarrow b=\frac{1}{2}; c=a=\frac{1}{4}\)