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

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

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

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

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

\(D=-2\)

Answer 2

CÂU THỨ 2 NHA !!!!!!

XÉT:        \(2VT=2a\sqrt{b-1}+2b\sqrt{a-1}\)

=>    \(2VT=a.2.\sqrt{1}.\sqrt{b-1}+b.2.\sqrt{1}.\sqrt{a-1}\)

TA ÁP DỤNG BĐT CAUCHY 2 SỐ SẼ ĐƯỢC: 

=>    \(2VT\le a\left(1+b-1\right)+b\left(1+a-1\right)\)

=>   \(2VT\le ab+ab\)

=>   \(2VT\le2ab\)

=>   \(VT\le ab\)

=> TA CÓ ĐIỀU PHẢI CHỨNG MINH.

Answer 3

\(D=\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}-\sqrt{2+2\sqrt{2}+1}\)

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

Answer 4

câu 2 (cách khác)

ta có (**) \(2\sqrt{ab}\le a+b\Leftrightarrow0\le\left(\sqrt{a}-\sqrt{b}\right)^2\)(bđt đúng với a>=0; b>=0)

áp dụng bđt (**) 

\(\sqrt{b-1}=\sqrt{1\left(b-1\right)}\le\frac{b-1+1}{2}=\frac{b}{2}\)(do b-1>=0)

\(\Rightarrow a\sqrt{b-1}\le\frac{ab}{2}\)(do b>0; a-1>=0)

vậy \(a\sqrt{b-1}+b\sqrt{a-1}\le\frac{ab}{2}+\frac{ab}{2}=ab\)