Question

两道高一关于基本不等式的数学题

1、已知:x^2+y^2=a,m^2+n^2=b(a,b>0),求mx+ny最大值. 2、设a，b属于R+，求证：a/√b + b/√a ≥√a + √b.

Answer 1

1.柯西不等式二维形式

http://baike.baidu.com/link?url=yxfuBCK80JLHS7u-aIcVpX2jS1a5IPfbOE4J25B6eQZb7Jsh4xU2_MPkNlBgjBRQ#1
ab=(x^2+y^2)(m^2+n^2)≥(mx+ny)^2
当且仅当my=bx时等号成立
所以(mx+ny)min=√ab

2.(a/√b +b/√a)-√a-√b
=(a/√b -√b)+(b/√a -√a)
通分，得
=(a-b)/√b +(b-a)/√a
=(a-b)/√b -(a-b)/√a
=(a-b)[1/√b -1/√a]
=[(a-b)(√a -√b)]/√(ab) ≥0 注：这里若判断不出来可以分类讨论a大约等于b 或a＜b

Answer 2

1. 由sinα的平方+cosα的平方=1

   设x=√a*sinA  y=√a*sinB

      m=√b*cosB  y=√b*cosB

   则mx+ny=√ab*sin（A+B）小于等于√ab

2. a/√b+ √b≥2√a
   b/√a+ √a≥2√b
   于是a/√b+ b/√a+√a+ √b≥2√a+2√b
    a/√b+ b/√a≥√a+√b

Answer 3

x^2+y^2+m^2+n^2=a+b
x^2+m^2>=2mx;y^2+n^2>=2ny
a+b>=2mx+2ny
得mx+ny<=a+b/2

Answer 4

1. (mx+ny)^2≤（x^2+y^2）（m^2+n^2）=ab
   

   mx+ny≤根号下ab

2. 同上 左边=（√a）^2 /√b + (√b)^2/√a≥ （√a+√b）^2/ （√a+√b）=√a + √b
   
   
   
   

Answer 5