Diseguaglianza a + b + c = 3
Diseguaglianza a + b + c = 3
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
$
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darkcrystal
- Messaggi: 706
- Iscritto il: 14 set 2005, 11:39
- Località: Chiavari
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
Ultima modifica di darkcrystal il 11 ott 2007, 17:10, modificato 1 volta in totale.
"Solo due cose sono infinite: l'universo e la stupidità dell'uomo, e non sono tanto sicuro della prima" - Einstein
Membro dell'EATO
Membro dell'EATO
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 $$ 27-x^3-27x+9x^2 \ge 12-4x^3 $
$ 3x^3+9x^2-27x+15 \ge 0 $ $ \longrightarrow $ $ x^3+3x^2-9x+5 \ge 0 $. This polynomial is 0 for x=-5. The other 2 roots are complex roots, so in R we have as minimum x=-5.
I hope this solution is correct...
Now let's try with x<-1.
We have this solution (a,b,x). $ a+b+x=3 $ $ a+b=3-x $.
We havethe minimum value for $ a^3+b^3 $ that $ a+b=3-x $ when a=b.
So let's try with (a,a,x). We have that $ a^3+a^3 \ge 3-x^3 $.
but $ a=\frac {3-x}{2} $ so$ (3-x)^3 \ge 12-4x^3 $ $ \longrightarrow $$ 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...
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.
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.
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.
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.