If $abc(a+b+c)=3$, prove that $(a+b)(b+c)(c+a)\geq 8$
My work. $$abc(a+b+c)=3$$ $$a^2bc+ab^2c+abc^2=3$$ Adding $5$ to both sides $$a^2bc+ab^2c+abc^2+5=5+3$$ $$a^2bc+ab^2c+abc^2+5=8$$ . . ... .
Asked
Active
Viewed 84 times
7
-
are the variables assumed to be positive? – Dr. Sonnhard Graubner Apr 14 '17 at 10:58
-
@Dr.Sonnhard Graubner, I don't know. That's not stated in the question. – Aryabhatta Apr 14 '17 at 11:03
-
3@Dr.SonnhardGraubner I think so. Consider $a=b=-1,c=3$. – Yuxiao Xie Apr 14 '17 at 11:06