已知a,b为正整数,x,y>0且a/x+b/y=1,求证(x+y)≥(√a+√b)^2
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/06 22:18:34
![已知a,b为正整数,x,y>0且a/x+b/y=1,求证(x+y)≥(√a+√b)^2](/uploads/image/z/12594653-53-3.jpg?t=%E5%B7%B2%E7%9F%A5a%2Cb%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%2Cx%2Cy%3E0%E4%B8%94a%2Fx%2Bb%2Fy%3D1%2C%E6%B1%82%E8%AF%81%28x%2By%29%E2%89%A5%28%E2%88%9Aa%2B%E2%88%9Ab%29%5E2)
xN0_cR[r9 PE[a@ *C*E"XB( 5-Kk;
8N
jUX9.|QDRLMzHC<]gErjy -aQaO8%;V* _AV3cc
<ĉ){~)=KrO8)4`\vM4YMx\M"(;xej(Dz=Z{&k(e|{fU~\A3? 6
已知a,b为正整数,x,y>0且a/x+b/y=1,求证(x+y)≥(√a+√b)^2
已知a,b为正整数,x,y>0且a/x+b/y=1,求证(x+y)≥(√a+√b)^2
已知a,b为正整数,x,y>0且a/x+b/y=1,求证(x+y)≥(√a+√b)^2
已知a,b为正整数,x,y>0且a/x+b/y=1,
则 (x+y)=(x+y)*(a/x+b/y)=a+b+ay/x+bx/y>=a+b+2根号(ab)=(√a+√b)^2
x+y
=(x+y)(a/x+b/y)
=a+b+ay/x+bx/y
≥a+b+2√(ab)
=(√a+√b)^2
解由柯西不等式1=a/x+b/y》(√a+√b)^2/x+y
所以(x+y)≥(√a+√b)^2
事实上a/x+b/y=(a/x+b/y)*(x+y)/(x+y),对分子用柯西不等式即得,》(√a+√b)^2/x+y