设x,y,z为正实数, 求证
√[x/(x+y)]+√[y/(y+z)]+√[z/(z+x)]≤(3√2)/2

设x,y,z为正实数, 求证
√[x/(x+y)]+√[y/(y+z)]+√[z/(z+x)]≤(3√2)/2 (1)
证明 设a=y+z, b=z+x, c=x+y，则可解得:
x=(b+c-a)/2, y=(c+a-b)/2, z=(a+b-c)/2。
对所证不等式作置换得:
T=√[(b+c-a)/(2c)]+√[(c+a-b)/(2a)]+√[(a+b-c)/(2b)]≤(3√2)/2 (2)
显然a,b,c三线段可构成一个三角形，设a,b,c三边对应的三角形三内角分别为A,B,C。则由正弦定理化简得:
(b+c-a)/(2c)=cos(A/2)*sin(B/2)/cos(C/2); (3-1)
(c+a-b)/(2a)=cos(B/2)*sin(C/2)/cos(A/2); (3-2)
(a+b-c)/(2b)=cos(C/2)*sin(A/2)/cos(B/2). (3-3)
将上述三式代入(2)式得:
T=√[cos(A/2)*sin(B/2)/cos(C/2)]+√[cos(B/2)*sin(C/2)/cos(A/2)]+√[cos(C/2)*sin(A/2)/cos(B/2)]
=[cos(A/2)*√sinB+cos(B/2)*√sinC+cos(C/2)*√sinA]/√[2cos(A/2)*cos(B/2)*cos(C/2)]
运用Cauchy不等式得:
T≤√{[cos^2(A/2)+cos^2(B/2)+cos^2(C/2)]*(sinA+sinB+sinC)}/√[2cos(A/2)*cos(B/2)*cos(C/2)]
再将已知不等式:
[cos(A/2)]^2+[cos(B/2)]^2+[cos(C/2)]^2≤9/4 (4)
和已知恒等式:
sinA+sinB+sinC=4cos(A/2)*cos(B/2)*cos(C/2) (5)
一并代入得:
T≤√2[cos^2(A/2)+cos^2(B/2)+cos^2(C/2)]≤(3√2)/2. 证毕。