Đề bài cho dư. Chỉ cần vế dưới là đủ rồi.
Điều kiện: \(x,y\ge1\)
Ta có:
\(\sqrt{x\left(xy-x\right)}+\sqrt{y\left(xy-y\right)}\le\dfrac{x+xy-x}{2}+\dfrac{y+xy-y}{2}=xy\)
Dấu = xảy ra khi \(x=y=2\)
PT(1)⇔(x+2013−−−−−−−√−x+2012−−−−−−−√)+(x+2012−−−−−−−√−x+2011−−−−−−−√)PT(1)⇔(x+2013−x+2012)+(x+2012−x+2011)
=(y+2012−−−−−−−√−y+2011−−−−−−−√)+(z+2013−−−−−−−√−z+2012−−−−−−−√)=(y+2012−y+2011)+(z+2013−z+2012)
⇔1x+2012−−−−−−−√+x+2011−−−−−−−√+1 sqrtx+2013+x+2012−−−−−−−√⇔1x+2012+x+2011+1 sqrtx+2013+x+2012
=1y+2012−−−−−−−√+y+2011−−−−−−−√+1 sqrtz+2013+z+2012−−−−−−−√=1y+2012+y+2011+1 sqrtz+2013+z+2012
Giả sử: x≥y;x≥z, ta có:
1x+2012−−−−−−−√+x+2011−−−−−−−√≤1y+2012−−−−−−−√+y+2011−−−−−−−√1x+2012+x+2011≤1y+2012+y+2011
1x+2013−−−−−−−√+x+2012−−−−−−−√≤1z+2013−−−−−−−√+z+2012−−−−−−−√1x+2013+x+2012≤1z+2013+z+2012
⇒1x+2012−−−−−−−√+x+2011−−−−−−−√+1 sqrtx+2013+x+2012−−−−−−−√⇒1x+2012+x+2011+1 sqrtx+2013+x+2012
≤1y+2012−−−−−−−√+y+2011−−−−−−−√+1 sqrtz+2013+z+2012−−−−−−−√≤1y+2012+y+2011+1 sqrtz+2013+z+2012
Mà 1x+2012−−−−−−−√+x+2011−−−−−−−√+1x+2013−−−−−−−√+x+2012−−−−−−−√1x+2012+x+2011+1x+2013+x+2012
=1y+2012−−−−−−−√+y+2011−−−−−−−√+1 sqrtz+2013+z+2012−−−−−−−√=1y+2012+y+2011+1 sqrtz+2013+z+2012
⇒x=y=z⇒x=y=z
Đây là đề thi trên báo THTT phải không
