# Inequality, not very hard but nice olympiad problem

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• Sep 1st 2007, 07:32 AM
Ununuquantium
Inequality, not very hard but nice olympiad problem
Find the smallest real number $\displaystyle x$, such that: For arbitrary real numbers $\displaystyle a, b, c\geqslant x$ and $\displaystyle a + b + c = 3$ this inequality is correct: $\displaystyle a^{3} + b^{3} + c^{3}\geqslant 3$
• Sep 5th 2007, 03:37 AM
mahurshi
Smallest real number x where a,b,c >=x, a+b+c=3, a^3+b^3+c^3>=3
Quote:

Originally Posted by Ununuquantium
Find the smallest real number $\displaystyle x$, such that: For arbitrary real numbers $\displaystyle a, b, c\geqslant x$ and $\displaystyle a + b + c = 3$ this inequality is correct: $\displaystyle a^{3} + b^{3} + c^{3}\geqslant 3$

Is the answer $\displaystyle x = 0$ ? (where $\displaystyle a = b = c = 1$)

I got this by trial and error, but I am still half asleep right now. :)

Mahurshi Akilla
• Sep 5th 2007, 04:07 AM
janvdl
Quote:

Originally Posted by Ununuquantium
Find the smallest real number $\displaystyle x$, such that: For arbitrary real numbers $\displaystyle a, b, c\geqslant x$ and $\displaystyle a + b + c = 3$ this inequality is correct: $\displaystyle a^{3} + b^{3} + c^{3}\geqslant 3$

I would say $\displaystyle x = 1$, but i'm not sure how to prove it mathematically.
• Sep 5th 2007, 06:46 AM
Jhevon
Quote:

Originally Posted by mahurshi
Is the answer $\displaystyle x = 0$ ? (where $\displaystyle a = b = c = 1$)

I got this by trial and error, but I am still half asleep right now. :)

Mahurshi Akilla

what about a = b = -1, c = 5

then we could have x = -1

Quote:

Originally Posted by janvdl
I would say $\displaystyle x = 1$, but i'm not sure how to prove it mathematically.

what about a = b = 0 and c = 3

then we could have x = 0

Quote:

Originally Posted by Ununuquantium
Find the smallest real number $\displaystyle x$, such that: For arbitrary real numbers $\displaystyle a, b, c\geqslant x$ and $\displaystyle a + b + c = 3$ this inequality is correct: $\displaystyle a^{3} + b^{3} + c^{3}\geqslant 3$

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 = $\displaystyle \sqrt {2}$, b = $\displaystyle - \sqrt {2}$, c = 3. those also fulfil the conditions given. so we could write x = $\displaystyle - \sqrt {2}$.

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 = $\displaystyle \infty$, b = $\displaystyle - \infty$, c = 3

then we could say x = $\displaystyle -\infty$, 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 $\displaystyle \infty - \infty$ 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?
• Sep 5th 2007, 07:14 AM
Ununuquantium
I think $\displaystyle x=-5$ - but I am not sure and I cannot prove it...
• Sep 5th 2007, 08:27 AM
Jhevon
Quote:

Originally Posted by Ununuquantium
I think $\displaystyle x=-5$ - but I am not sure and I cannot prove it...

did you read my post above? did you have a problem with it? are you sure you have the question right? as it stands, x can be anything i believe
• Sep 5th 2007, 08:33 AM
Ununuquantium
No, the question is formed absolutely properly. Yor answer is wrong...
• Sep 5th 2007, 08:42 AM
Jhevon
Quote:

Originally Posted by Ununuquantium
No, the question is formed absolutely properly. Yor answer is wrong...

you're sure the answer is around -5?
• Sep 5th 2007, 08:54 AM
Ununuquantium
I am not absolutely sure, but I think that the answer is $\displaystyle 5$
• Sep 5th 2007, 09:15 AM
Jhevon
Quote:

Originally Posted by Ununuquantium
Find the smallest real number $\displaystyle x$, such that: For arbitrary real numbers $\displaystyle a, b, c\geqslant x$ and $\displaystyle a + b + c = 3$ this inequality is correct: $\displaystyle a^{3} + b^{3} + c^{3}\geqslant 3$

Quote:

Originally Posted by Ununuquantium
I am not absolutely sure, but I think that the answer is $\displaystyle 5$

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

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
• Sep 5th 2007, 09:40 AM
Ununuquantium
Ugh..... I cannot get through that problem...
• Sep 5th 2007, 12:48 PM
topsquark
Quote:

Originally Posted by Jhevon
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

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

Gee, Jhevon, are you sure it isn't x = -50? :D

-Dan
• Sep 5th 2007, 03:48 PM
mahurshi
Quote:

Originally Posted by Jhevon
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

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

you're right. $\displaystyle x$ can be a number that's much smaller too. i didn't realize that when i posted earlier.
• Sep 5th 2007, 05:32 PM
Jhevon
Quote:

Originally Posted by topsquark
Gee, Jhevon, are you sure it isn't x = -50? :D

-Dan

of course i was being sarcastic here :p

as i said, any x you choose, i choose a smaller x and still find real numbers a,b, and c that fulfill all conditions
• Oct 15th 2007, 11:17 AM
terafull
I think that, "for arbitrary" in the contents means "for any(every)"
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