(x^2+y^2)≥[(x+y)^2]/2
√(x^2+y^2)≥(√2/2)*(x+y)同理
√(y^2+z^2)≥(√2/2)*(z+y),√(z^2+x^2)≥(√2/2)*(z+x)
√(x^2+y^2)+√(y^2+z^2)+√(z^2+x^2)≥(√2/2)*(x+y)+(√2/2)*(z+y)+(√2/2)*(z+x)=√2*(x+y+z)
等式成立条件x=y=z
急.求证:√(x^2+y^2)+√(y^2+z^2)+√(z^2+x^2)≥√2*(x+y+z)
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