Inequality with Min. value.

[man111]

If $ a , b ,c$ are distinct positive integers such that $ab + bc + ca\geq 107$ , then find the minimum value of $\frac{1}{6}( a^3 + b^3 + c^3 - 3abc )$

[Руст]

Because $a^3+b^3+c^3\ge 3abc$ for positive numbers $a,b,c$ minimal value of $a^3+b^3+c^3-3abc$ when $a=b=c$. When $a,b,c$ are distinct integers vbnimal mean, when $b=a+1,c=a-1$ and
$f=a^3+b^3+c^3-3abc=9a, ab+bc+ca=3a^2-1\ge 107\to a\ge 6$.
Therefore $f=54, (a,b,c)=(5,6,7).$

[man111]

Thanks.

[man111]

man111 в сообщении #396665 писал(а):

Thanks. Pyct for Nice solution.