Find the smallest real number , such that: For arbitrary real numbers and this inequality is correct:
then we could have x = -1
then we could have x = 0
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?
I claim the answer is x = -25
To show this works, I choose a = -25, b = 25, c = 3
is a,b,c >= x? yes! -25, 25 and 3 are all greater than or equal to -25. so the first condition is fulfilled.
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:
is the inequality a^3 + b^3 + c^3 >= 3 correct? yes!
(-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