Question

正实数a1 a2 满足 [a1]^2+[a2]^2=1,会有 a1+a2≤√2.若 n 个正实数 [a1]^2+[a2]

Answer 1

用柯西不等式,

（a1+a2+a3 + .. +an）^2

≤ ([a1]^2+[a2]^2+[a3]^2 + .. +[an]^2)(1^2+1^2+ .. +1^2)

= n([a1]^2+[a2]^2+[a3]^2 + .. +[an]^2)=n

所以 a1+a2+a3 + .. +an≤√n , 等号当且仅当 a1=a2=a3= .. =an.