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设a,b,c∈R+,求证:a2/b+b2/c+c2/a≥a+b+c若a、b∈R+,求证：a/根号b+b/根号a≥根号a+根号b
题目详情
设a,b,c∈R+,求证:a2/b+b2/c+c2/a≥a+b+c
若a、b∈R+,求证：a/根号b+b/根号a≥根号a+根号b
若a、b∈R+,求证：a/根号b+b/根号a≥根号a+根号b
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答案和解析
a²/b+b²/c+c²/a-(a+b+c)
=(a²-b²)/b+(b²-c²)/c+(c²-a²)/a
≥(a²-b²)/(a+b+c)+(b²-c²)/(a+b+c)+(c²-a²)/(a+b+c)
=0
所以a²/b+b²/c+c²/a≥a+b+c
(a/√b)+(b/√a)-√a-√b
=[(a-b)/√b]+[(b-a)/√a]
≥[(a-b)/(√a+√b)]+[(b-a)/(√a+√b)]
=0
所以(a/√b)+(b/√a)≥√a+√b
=(a²-b²)/b+(b²-c²)/c+(c²-a²)/a
≥(a²-b²)/(a+b+c)+(b²-c²)/(a+b+c)+(c²-a²)/(a+b+c)
=0
所以a²/b+b²/c+c²/a≥a+b+c
(a/√b)+(b/√a)-√a-√b
=[(a-b)/√b]+[(b-a)/√a]
≥[(a-b)/(√a+√b)]+[(b-a)/(√a+√b)]
=0
所以(a/√b)+(b/√a)≥√a+√b
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