均值不等式,x,y,z都是正实数,有x^2+(y^2)/2≥xy√2.....①(等号成立x^2=(y^2)/2(y^2)/2+z^2≥yz√2.....②(等号成立(y^2)/2=z^2①+②得x^2+y^2/2+y^2/2+z^2≥xy√2+yz√2=√2(xy+yz)所以(xy+yz)/(x^2+y^2+z^2)≤1/√2=(√2)/2故当且仅当x^2=(y^2)/2=z^2,即x=(√2)y/2=z时,(xy+yz)/(x^2+y^2+z^2)取得最大值(√2)/2