The minimum value of the expression $(a+b)(b+c)$, under the constraint $abc(a+b+c)=1$ with positive real numbers $a$, $b$, and $c$, is established through mathematical proof. Participants in the discussion suggest specific values such as $a=1$, $b=-1+\sqrt{2}$, and $c=1$ to explore the conditions of the problem. The discussion emphasizes the importance of understanding the relationships between the variables to derive the minimum effectively.
PREREQUISITESMathematicians, students studying optimization techniques, and anyone interested in algebraic proofs and inequalities.
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]