The discussion revolves around finding the minimum value of the expression $(a+b)(b+c)$ under the constraint that $a, b,$ and $c$ are positive real numbers satisfying the condition $abc(a+b+c)=1$. The focus is on providing a proof for this minimum value.
The discussion appears to have unresolved points, particularly regarding the validity of the suggested values for $a$, $b$, and $c$, and whether they meet the problem's constraints. There is no consensus on the approach or solution presented.
Participants have not fully addressed the implications of the positivity constraint on the values of $a$, $b$, and $c$. The repeated suggestion of negative values raises questions about the applicability of those examples.
anemone said:Find, with proof, the minimum value of $(a+b)(b+c)$ where $a,\,b$ and $c$ are positive real numbers satisfying the condition $abc(a+b+c)=1$.
greg1313 said:$$(a+b)(b+c)=ab+ac+bc+b^2=ac+b(a+b+c)=ac+\dfrac{1}{ac}$$
AM-GM:
$$\dfrac{ac+\dfrac{1}{ac}}{2}\ge\sqrt{ac\cdot\dfrac{1}{ac}}=1$$
$$\implies\min\left[(a+b)(b+c)\right]=2$$
greg1313 said:$$(a+b)(b+c)=ab+ac+bc+b^2=ac+b(a+b+c)=ac+\dfrac{1}{ac}$$
AM-GM:
$$\dfrac{ac+\dfrac{1}{ac}}{2}\ge\sqrt{ac\cdot\dfrac{1}{ac}}=1$$
$$\implies\min\left[(a+b)(b+c)\right]=2$$
kaliprasad said:It has been shown that $(a+b)(b+c) >=2$ but I think it needs to be shown that $(a+b)(b+c) = 2 $ is met for at least one set of a,b,c
I am not telling that you are wrong. what I am telling is that I do not get it
greg1313 said:[sp]I'm not clear on what it is you don't understand. However, try $a=1,b=-1+\sqrt2,c=1$.[/sp]