Find the smallest real number, such that: For arbitrary real numbersandthis inequality is correct:

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- September 1st 2007, 07:32 AMUnunuquantiumInequality, not very hard but nice olympiad problem
**Find the smallest real number****, such that: For arbitrary real numbers****and****this inequality is correct:** - September 5th 2007, 03:37 AMmahurshiSmallest real number x where a,b,c >=x, a+b+c=3, a^3+b^3+c^3>=3
- September 5th 2007, 04:07 AMjanvdl
- September 5th 2007, 06:46 AMJhevon
what about a = b = -1, c = 5

then we could have x = -1

what about a = b = 0 and c = 3

then we could have x = 0

this is a weird question. perhaps it should read "the**greatest**real number x," or "the smallest**nonnegative**real number x," since we can probably find an infinite number of combinations of numbers that fulfil the required conditions, especially since we don't have to restrict ourselves to integers (i can't prove this, but i think it makes sense). Consider even, a = , b = , c = 3. those also fulfil the conditions given. so we could write x = .

In fact, (without loss of generallity) if a + b = 0, and c = 3, say, then we fulfil these conditions for an infinite set of numbers, since there are infinite possibilities for a and b such that a + b = 0. i think we can show that by induction. but let's just consider extremes here:

what about a = , b = , c = 3

then we could say x = , so whatever x you tell me, i can tell you a smaller one for which it works

(of course i realize that no number can equal infinity, and that may not necessarilly be zero--or is it? but i just want to show we can get as small a number as we want for x)

i could say x = -500000, then choose a = 500000, b = -500000 and c = 3, and those work!

ok, enough rambling, i think you guys get my point, or am i not making sense at all? - September 5th 2007, 07:14 AMUnunuquantium
I think - but I am not sure and I cannot prove it...

- September 5th 2007, 08:27 AMJhevon
- September 5th 2007, 08:33 AMUnunuquantium
No, the question is formed

**absolutely**properly. Yor answer is wrong... - September 5th 2007, 08:42 AMJhevon
- September 5th 2007, 08:54 AMUnunuquantium
I am not absolutely sure, but I think that the answer is

- September 5th 2007, 09:15 AMJhevon
ok. let's say you made a huge mistake, and the answer was actually even smaller, say -25.

I claim the answer is x = -25

To show this works, I choose a = -25, b = 25, c = 3

yes! -25, 25 and 3 are all greater than or equal to -25. so the first condition is fulfilled.

is a,b,c >= x?

**is a + b + c = 3?**yes! -25 + 25 + 3 = 3. so our second condition is fulfilled.

**are a,b,c, and x real numbers?**yes! they are integers. so that condition is fulfilled

Now the moment of truth:

yes!

is the inequality a^3 + b^3 + c^3 >= 3 correct?

(-25)^3 + (25)^3 + 3^3 = -(25)^3 + (25)^3 + 3^3 = 27 >= 3

so all conditions hold for x = -25. so x = -25 is the answer - September 5th 2007, 09:40 AMUnunuquantium
Ugh..... I cannot get through that problem...

- September 5th 2007, 12:48 PMtopsquark
- September 5th 2007, 03:48 PMmahurshi
- September 5th 2007, 05:32 PMJhevon
- October 15th 2007, 11:17 AMterafull
I think that, "for arbitrary" in the contents means "for any(every)"