Maximum value of $abc$ for $a, b, c > 0$ and $ab + bc + ca = 12$

Question

$a,b,c$ are three positive real numbers such that $ab+bc+ca=12$.

Then find the maximum value of $abc$

Answer 1

Applying the inequality $AM \ge GM $ on {$ab,bc,ca$}, we get: $$\frac{ab+bc+ca}{3} \ge (a^2.b^2.c^2)^{\frac13}$$ $\Rightarrow$ $(abc)^{\frac23} \le \frac{12}{3} = 4$

i.e. the maximum value of abc is $4^{\frac32}$ = $8$.

Also, maximum value of $8$ is attained for $a=b=c=2$

Answer 2

Or, the same answer also comes by intuition. since a,b,c are positive numbers, $(ab).(bc).(ca)$ is maximum only when $ab=bc=ca=\frac{12}{3}=4$. therefore, $ abc={((ab)(bc)(ca))}^\frac{1}{2}=4^\frac{3}{2}=8$.

Answer 3

I do not understand that. Same for any number can make this combination.

equation: $XY+XZ+YZ=N$

Solutions in integers can be written by expanding the number of factorization: $N=ab$

And vospolzovavschis solutions of Pell's equation: $p^2-(4k^2+1)s^2=1$

$k$ - what some integer number given by us.

Solutions can be written:

$X=ap^2+2(ak+b+a)ps+(2(a-2b)k+2b+a)s^2$

$Y=2(ak-b)ps+2(2ak^2+(a+2b)k+b)s^2$

$Z=bp^2-2(2b+a)kps+(4bk^2-2ak-b)s^2$

And more:

$X=-2bp^2+2(k(4b+a)+b)ps-2((4b+2a)k^2+(2b-a)k)s^2$

$Y=-(2b+a)p^2+2(k(4b+a)-b-a)ps-(8bk^2-(4b+2a)k+a)s^2$

$Z=bp^2-2(2b+a)kps+(4bk^2-2ak-b)s^2$

Perhaps these formulas for someone too complicated.

Then equation: $XY+XZ+YZ=N$

If we ask what ever number: $p$

That the following sum can always be factored: $p^2+N=ks$

Solutions can be written.

$X=p$

$Y=s-p$

$Z=k-p$

8 - really is the maximum number of turns. But the smallest one can hardly get.