Help me, please !

G

Guest

Guest
There are 3 real numbers a,b,c (a,b,c>0).

We know that : ab+bc+ca=3

Proove that a^3+b^3+c^3 + 6abc >= 9
 
asa starting point look to expand your statment of ab+bc+ca=3
by cubing it

(ab+bc+ca) ^ 3 = 27

and or sqaring it

(ab+bc+ca) ^2 = 9
 
Isn't there a chance of introducing extraneous roots into the solution if Brolly chooses to square both sides of the equation?
 
Roots have nothing to do with proving inequalities.
At least not in this case.
 
Do you know the distinction between PROVE and SOLVE?
 
There are many interesting relationships here.
You can see that the first four terms of the cube are your ‘left hand side’.
The remaining terms have an interesting relationship.
\(\displaystyle (a + b + c)^3 = a^3 + b^3 + c^3 + 6abc + 3a^2 b + 3a^2 c + 3b^2 a + 3b^2 c + 3c^2 a + 3c^2 b\)

\(\displaystyle 3(a + b)(a + c)(b + c) = 6abc + 3a^2 b + 3a^2 c + 3b^2 a + 3b^2 c + 3c^2 a + 3c^2 b\)

BUT I have not been able to go beyond that.
There may be a trick in writing ab+ac+bc that I just do not see.
Another possibility is a misprint.
Double-check you posting. Is it correct?
 
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