Question

cho a,b>1. chứng minh rằng

\(\dfrac{6}{a\sqrt{b-1}+b\sqrt{a-1}}+\sqrt{3ab+4}\ge\dfrac{11}{2}\)

Answer 1

Áp dụng AM-GM: \(a\sqrt{b-1}+b\sqrt{a-1}\le a.\dfrac{b-1+1}{2}+b.\dfrac{a-1+1}{2}=ab\)

\(VT\ge\dfrac{6}{ab}+\sqrt{3ab+4}\)

( dự đoán dấu = xảy ra khi a=b=2)

Áp dụng cauchy-schwarz:

\(\dfrac{6}{ab}=\dfrac{18}{3ab}+\dfrac{2}{4}-\dfrac{1}{2}\ge\dfrac{\left(\sqrt{18}+\sqrt{2}\right)^2}{3ab+4}-\dfrac{1}{2}=\dfrac{32}{3ab+4}-\dfrac{1}{2}\)

Áp dụng AM-GM một lần nữa:

\(VT\ge\dfrac{32}{3ab+4}+\sqrt{3ab+4}-\dfrac{1}{2}=\dfrac{32}{3ab+4}+\dfrac{\sqrt{3ab+4}}{2}+\dfrac{\sqrt{3ab+4}}{2}-\dfrac{1}{2}\ge3\sqrt[3]{\dfrac{32}{4}}-\dfrac{1}{2}=\dfrac{11}{2}\)

Dấu = xảy ra khi a=b=2

P/s: Nothing