Question

\(a\sqrt{b-1}+b\sqrt{a-1}\le ab\)

Answer 1

ĐKXĐ: \(a\ge1;b\ge1\)

BĐT đã cho tương đương:

\(\dfrac{a\sqrt{b-1}}{ab}+\dfrac{b\sqrt{a-1}}{ab}\le1\Leftrightarrow\dfrac{\sqrt{b-1}}{b}+\dfrac{\sqrt{a-1}}{a}\le1\)

Ta có: \(\dfrac{1.\sqrt{b-1}}{b}+\dfrac{1.\sqrt{a-1}}{a}\le\dfrac{1+b-1}{2b}+\dfrac{1+a-1}{2a}=\dfrac{1}{2}+\dfrac{1}{2}=1\) (đpcm)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{a-1}=1\\\sqrt{b-1}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=2\end{matrix}\right.\)