【注：當(dāng)x,y≥0時(shí),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.】證明：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≥0.兩式相乘得[(x+y)/2]×√[2(x+y)]≥(√x+√y)√(xy)=x√y+y√x.綜上可知,[(x+y)²/2]+[(x+y)/4]≥x√y+y√x.等號(hào)僅當(dāng)x=y=0時(shí)取得.