已知我x>=0,y>=0,求证:1/2(x+y)^2+1/4(x+y)>=x根号下y+y根号下x

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/05 17:25:48
已知我x>=0,y>=0,求证:1/2(x+y)^2+1/4(x+y)>=x根号下y+y根号下x
xRJAe]|1% fI $K"z 1AW^ݙ_&Z =2{sϹ3KUwX }G$0w,(tWF!X;coT׼xOӖ|X*ݔj qX3Y~5|ȩ1[&a98r]Gr0 5\CBdu*Uvu b ̪$))bMۃzF.Q+fq̒PW2WNbEkHh?`ܔoфq_ \J,¤ߞ _^z ش;}juf

已知我x>=0,y>=0,求证:1/2(x+y)^2+1/4(x+y)>=x根号下y+y根号下x
已知我x>=0,y>=0,求证:1/2(x+y)^2+1/4(x+y)>=x根号下y+y根号下x

已知我x>=0,y>=0,求证:1/2(x+y)^2+1/4(x+y)>=x根号下y+y根号下x
【注:当x,y≥0时,由均值不等式知,x+y≥2√(xy).===>2(x+y)≥x+2√(xy)+y=(√x+√y)²,===>√[2(x+y)]≥√x+√y.即有:x+y≥2√(xy),且√[2(x+y)]≥√x+√y.】证明:因x,y≥0,故x+y≥0.(1).由均值不等式可知,[(x+y)²/2]+[(x+y)/4]≥2√{[(x+y)²/2]×[(x+y)/4]}=[(x+y)/2]×√[2(x+y)].(2)又(x+y)/2≥√(xy)≥0,√[2(x+y)]≥√x+√y≥0.两式相乘得[(x+y)/2]×√[2(x+y)]≥(√x+√y)√(xy)=x√y+y√x.综上可知,[(x+y)²/2]+[(x+y)/4]≥x√y+y√x.等号仅当x=y=0时取得.