(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
所以,結(jié)論成立.
1/(√3+√2)-(√5-2)
=(√3-√2)-(√5-2)
=(√3+√4)-(√2+√5)
(√3+√4)^2=7+2√12
(√2+√5)^2=7+2√10
(√3+√4)>(√2+√5)
1/(√3+√2)-(√5-2) >0
1/(√3+√2)>√5-2
要證明ax^2+by^2 >= (ax+by)^2
即證明ax^2+by^2 - (ax+by)^2 >= 0
ax^2+by^2-(ax+by)^2
= ax^2+by^2-(ax)^2-2abxy-(by)^2
= a(1-a)x^2+b(1-b)y^2-2abxy
根據(jù)已知a+b=1
= abx^2+aby^2-2abxy
= ab(x^2-2xy+y^2)
利用完全平方公式
= ab(x-y)^2
∵a,b都是正數(shù),且(x-y)^2 >= 0
∴ab(x-y)^2 >= 0
∴ax^2+by^2 - (ax+by)^2 >= 0成立
∴ax^2+by^2 >= (ax+by)^2成立