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標題:
Proof Inequality
發問:
Proof for every a, b, such that 0<=a<=b, one has a<=sqr(ab)<=(a+b)/2<=b.
最佳解答:
Consider (sqr(a) - sqr(b))^2 >= 0 a - 2 sqr(a) sqr(b) + b >= 0 (a+b)/2 > = sqr (ab) (a+b)/2 <= (b+b)/2 = b (since a<=b) sqr(ab) >= sqr(a*a) = a (since b>=a)
其他解答:
Proof Inequality
發問:
Proof for every a, b, such that 0<=a<=b, one has a<=sqr(ab)<=(a+b)/2<=b.
最佳解答:
Consider (sqr(a) - sqr(b))^2 >= 0 a - 2 sqr(a) sqr(b) + b >= 0 (a+b)/2 > = sqr (ab) (a+b)/2 <= (b+b)/2 = b (since a<=b) sqr(ab) >= sqr(a*a) = a (since b>=a)
其他解答:
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