Question

Cho \(D=\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)

CMR:Với mọi a,b>1 thì \(a\sqrt{b-1}+b\sqrt{a-1\le ab}\)

Answer 1

\(D=\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)

\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}\)

\(=\sqrt{2}-1-\sqrt{2}-1=-2\)

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Ta có: \(\left(\sqrt{a-1}-1\right)^2\ge0\forall a\ge1\)

\(\Leftrightarrow a-2\sqrt{a-1}\ge0\)

\(\Leftrightarrow\frac{\sqrt{a-1}}{a}\le\frac{1}{2}\)

Tương tự: \(\frac{\sqrt{b-1}}{b}\le\frac{1}{2}\)

\(\Rightarrow\frac{\sqrt{a-1}}{a}+\frac{\sqrt{b-1}}{b}\le1\)

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

Dấu "=" xảy ra \(\Leftrightarrow a=b=2\)