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標題:
inequality
發問:
最佳解答:
Obviously, p,q,r > 1 (1/p)+(1/q)+(1/r) obtains maximum when p,q and r are of the smallest possible values. However, they cannot be all 2 at the same time. So, when one of p,q and r is 2, the smallese possible value of the other two is 3. When one of them is 2 and one of them is 3: if the last one is 4, (1/2)+(1/3)+(1/4) = 13/12 ＞ 1 5, (1/2)+(1/3)+(1/5) = 31/30 ＞ 1 6, (1/2)+(1/3)+(1/4) = 1 7, (1/2)+(1/3)+(1/7) = 41/42＜ 1 For the last one ＞ 7, the value of (1/p)+(1/q)+(1/r) ＜ 41/42 Hence, 41/42 is the maximum possible value foe (1/p)+(1/q)+(1/r) if (1/p)+(1/q)+(1/r) ＜ 1. i.e. (1/p)+(1/q)+(1/r) ＜= 41/42 (I'm an S.4 student only. I don't know the formal proof)
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inequality
發問:
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Show that if p,q,r are positive integers satisfying (1/p)+(1/q)+(1/r) < 1 then (1/p)+(1/q)+(1/r) <= 41/42最佳解答:
Obviously, p,q,r > 1 (1/p)+(1/q)+(1/r) obtains maximum when p,q and r are of the smallest possible values. However, they cannot be all 2 at the same time. So, when one of p,q and r is 2, the smallese possible value of the other two is 3. When one of them is 2 and one of them is 3: if the last one is 4, (1/2)+(1/3)+(1/4) = 13/12 ＞ 1 5, (1/2)+(1/3)+(1/5) = 31/30 ＞ 1 6, (1/2)+(1/3)+(1/4) = 1 7, (1/2)+(1/3)+(1/7) = 41/42＜ 1 For the last one ＞ 7, the value of (1/p)+(1/q)+(1/r) ＜ 41/42 Hence, 41/42 is the maximum possible value foe (1/p)+(1/q)+(1/r) if (1/p)+(1/q)+(1/r) ＜ 1. i.e. (1/p)+(1/q)+(1/r) ＜= 41/42 (I'm an S.4 student only. I don't know the formal proof)
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