OliForum Matematico

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

x=0. AM=1 CM>1;
The inequality is $ 3*CM^3 \geqslant 3*AM $ that is obvious.

I know that the answer is -5 but can't prove it

If we choose $ (a,b,c) =(-y,0,y+3), y \geq 0 $, the inequality becomes $ -y^3+0+y^3+27+27y+9y^2 = 27+27y+9y^2 \geq 27 >3 $, so it would be $ x = - \infty $... it seems I can't get the problem statement...
If x had to be non-negative, then the correct answer would be Alex89's...

EDIT: I missed "arbitrary", I'm really sorry

Ok we know that for x=0 the inequality is correct. We also know that for -1<x<0 the inequality is also correct (it's easy because if $ -1 \le a \le 0 $then $ a^3 \ge a $ and we know that the bigger between a,b,c is $ \ge 3 $.

Now let's try with x<-1.

We have this solution (a,b,x). $ a+b+x=3 $ $ a+b=3-x $.
We havethe minimum value for $ a^3+b^3 $ that $ a+b=3-x $ when a=b.
So let's try with (a,a,x). We have that $ a^3+a^3 \ge 3-x^3 $.

but $ a=\frac {3-x}{2} $ so$ (3-x)^3 \ge 12-4x^3 $ $ \longrightarrow $$ 27-x^3-27x+9x^2 \ge 12-4x^3 $

$ 3x^3+9x^2-27x+15 \ge 0 $ $ \longrightarrow $ $ x^3+3x^2-9x+5 \ge 0 $. This polynomial is 0 for x=-5. The other 2 roots are complex roots, so in R we have as minimum x=-5.
I hope this solution is correct...

but why is a=b?? Could you explain?

Sorry but my solution is badly written... we're trying to know x. So for the smallest x we have a (a,b,x) solution. Now from $ a+b+x=3 $ we have $ a+b=3-x $ and we're trying to prove that $ a^3+b^3 \ge 3-x^3 $. Now let's take the minimum value for $ a^3+b^3 $. This value is for a=b (the minimum value is the one such as AM=CM).
So let's try with a=b. The solution is (a,a,x). We know that $ a=\frac {3-x}{2} $. Calculating we arrive at $ x^3+3x^2-9x+5 \ge 0 $ that is $ (x+5)(x-1)^2 \ge 0 $ that is verified for $ x \ge -5 $. The smallest x is -5.

Could you show how you've got a = b ?? I dont know what CM and AM are.

Sorry

We want to know when $ a^3+b^3 $ is minimum with $ a+b=3-x $.

This is CM-AM:
$ \sqrt[3]{\frac {a^3+b^3}{2}} \ge \frac{a+b}{2} $
and if a=b we have
$ \sqrt[3]{\frac {a^3+b^3}{2}}=\frac{a+b}{2} $ and $ a^3+b^3 $ is minimum. So we have to prove that the inequality of the problem is true if a=b , then for CM-AM we prove that the inequality for every a,b.