rewrite as $(a+b)(a+c)(b+c)\geq 4(ab+bc+ca)$ and multiply the LHS by the condition... one obtains it is equivalent to $(a+b)(a+c)+(b+c)(b+a)+(c+a)(c+b)\geq 4(ab+bc+ca)$ or equivalently $a^2+b^2+c^2\geq ab+bc+ca$...

Nice problem although it's a bit obvious \begin{align*}a+b+c-\frac{abc}{ab+bc+ca}&=\frac{(a+b)(b+)(c+a)}{ab+bc+ca} \\ &=\frac{a^2+b^2+c^2+3(ab+bc+ca)}{ab+bc+ca} \\ &\ge 4 \end{align*} P/s: Thanks for pointing out the silly mistake, tonypr

(Nice! I guess I'm too proud to not say I submitted this problem.. though never expected it to be Centro #3 ) Campos' and great math's solutions were pretty much what I submitted for the problem so I assume that that's the official solution as well. (@great math: Should be $a^2+b^2+c^2+3(ab+bc+ac)$ in the numerator, no? - however it still holds using the same solution)

great math wrote: Nice problem although it's a bit obvious \begin{align*}a+b+c-\frac{abc}{ab+bc+ca}&=\frac{(a+b)(b+)(c+a)}{ab+bc+ca} \\ &=\frac{a^2+b^2+c^2+3(ab+bc+ca)}{ab+bc+ca} \\ &\ge 4 \end{align*} P/s: Thanks for pointing out the silly mistake, tonypr And a small mistakes. $a+b+c-\frac{abc}{ab+bc+ca}=\frac{(a+b)(b+c)(c+a)}{ab+bc+ca} =\frac{a^2+b^2+c^2+3(ab+bc+ca)}{ab+bc+ca}$$\ge 4 $

;Given equality is equivalent to $\sum_{cyc}{a^2} + 3\sum_{cyc}{ab} = \sum_{sym}{a^2b} + 2abc.$ Apply Cauchy-Schwarz twice: $(\sum_{cyc}{(a+b)})(\sum_{cyc}{\frac{1}{a+b}}) \geq (\sum_{cyc}{1})^2,$ $\Rightarrow a+b+c \geq \frac{9}{2}.$ $(\sum_{cyc}{c^2(a+b)})(\sum_{cyc}{\frac{1}{a+b}}) \geq (\sum_{cyc}{c})^2,$ $\sum_{sym}{a^2b} \geq (a+b+c)^2,$ $\sum_{cyc}{a^2} + 3\sum_{cyc}{ab} - 2abc \geq (a+b+c)^2,$ $ab+bc+ca \geq 2abc,$ $\Rightarrow \frac{abc}{ab+bc+ca} \leq \frac{1}{2}.$ Putting the 2 inequalities with $\Rightarrow$ in front of them together, we get the desired result. It's a bit more complicated than the solutions given so far, but I thought that it illustrates a nice use of the CS inequality, and that sometimes going for a constant bound works (although it usually doesn't).

Notice that \[ a+b+c-\frac{abc}{ab+bc+ac}\ = \frac{(a+b)(a+c)(b+c)}{ab+bc+ac} \] but since \[ \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}=1 \] we have \[ \frac{(a+b)(a+c)(b+c)}{ab+bc+ac} = \frac{(a+b)(a+c) + (a+b)(b+c) + (a+c)(b+c) }{ab+bc+ac} \] Is just needed to proof \[ (a+b)(a+c) + (a+b)(b+c) + (a+c)(b+c)\ge 4(ab+bc+ac) \] wich is equivalent to \[ a^{2} + b^{2} +c^{2} \ge {ab+bc+ac} \] wich follows from rearrangement inequality.

what is Centroamerican?

iptgut wrote: what is Centroamerican? Centro American https://artofproblemsolving.com/community/c279h487096p4768162 $$(a+b+c)(ab+bc+ca)-abc=(a+b)(b+c)(c+a)=(a+b)(b+c)+(b+c)(c+a)+(c+a)(a+b)$$$$=a^2+b^2+c^2+3(ab+bc+ca)\ge 4(ab+bc+ca)$$