p>1,a,b>0时
a/(a+b)<[a/(a+b)]^(1/p)
b/(a+b)<[b/(a+b)]^(1/p)
所以,
1=a/(a+b)+b/(a+b)<[a/(a+b)]^(1/p)+[b/(a+b)]^(1/p)
即(a+b)^(1/p)其他时候此命题不见得对.