Find Minimum of $(a+b)(b+c)$ | Proof Provided

[anemone]

Find, with proof, the minimum value of $(a+b)(b+c)$ where $a,\,b$ and $c$ are positive real numbers satisfying the condition $abc(a+b+c)=1$.

 

[Greg]

Find, with proof, the minimum value of $(a+b)(b+c)$ where $a,\,b$ and $c$ are positive real numbers satisfying the condition $abc(a+b+c)=1$.

Spoiler

$$(a+b)(b+c)=ab+ac+bc+b^2=ac+b(a+b+c)=ac+\dfrac{1}{ac}$$

AM-GM:

$$\dfrac{ac+\dfrac{1}{ac}}{2}\ge\sqrt{ac\cdot\dfrac{1}{ac}}=1$$

$$\implies\min\left[(a+b)(b+c)\right]=2$$

 

[anemone]

Spoiler

$$(a+b)(b+c)=ab+ac+bc+b^2=ac+b(a+b+c)=ac+\dfrac{1}{ac}$$

AM-GM:

$$\dfrac{ac+\dfrac{1}{ac}}{2}\ge\sqrt{ac\cdot\dfrac{1}{ac}}=1$$

$$\implies\min\left[(a+b)(b+c)\right]=2$$

Awesome, greg1313! And thanks for participating!

 

[kaliprasad]

Spoiler

$$(a+b)(b+c)=ab+ac+bc+b^2=ac+b(a+b+c)=ac+\dfrac{1}{ac}$$

AM-GM:

$$\dfrac{ac+\dfrac{1}{ac}}{2}\ge\sqrt{ac\cdot\dfrac{1}{ac}}=1$$

$$\implies\min\left[(a+b)(b+c)\right]=2$$

Spoiler

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

 

[Greg]

Spoiler

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

[sp]I'm not clear on what it is you don't understand. However, try $a=1,b=-1+\sqrt2,c=1$.[/sp]

 

[kaliprasad]

[sp]I'm not clear on what it is you don't understand. However, try $a=1,b=-1+\sqrt2,c=1$.[/sp]

Spoiler

Now it is clear that there is a value shown above by you which satisfies the criteria.