Minimal value of the product $(x+y)(y+z)$

Question

Find the min. value of the product $\left(x + y\right)\left(y + z\right)$ given that $xyz\left(x + y + z\right) = 1$ and $x,y,z > 0$.

I want to solve this problem using geometry of a triangle:

• Let $a,b,c$ be the sides of the triangle and we can substitute $a = x + y\,,\,\, b = y + z\,,\,\, c = z + x$.
• $A^{2} = xyz\left(x + y + z\right)$ is just a squared area of a triangle and $\left(x + y\right)\left(y + z\right)$ is a part of another formula of triangle $$ \mbox{i.e.}\quad A = {1 \over 2}\,ab\sin\left(\theta\right)\quad \mbox{or}\quad A = {1 \over 2}\left(x + y\right)\left(y + z\right)\sin\left(\theta\right) $$ I' ve tried the same with other formulas but I don't know how to "extract" inequality.

Before the solution I would appreciate that you give a hint, and then a solution. Any help appreciated.

Answer 1

$$(x+y) (y+z) = 2\csc \theta$$

So what is the minimum value of $2\csc \theta$? What is the corresponding value of $\theta$? What kind of relationship does this indicate among $x, y, z$?

Answer 2

Let $a=x+y, \, b=y+z ,\, c = z+x, \, p = \frac{a+b+c}{2},$ then $$(x+y)(y+z) = ab,$$ and $$1 = \sqrt{xyz(x+y+z)} = \sqrt{p(p-a)(p-b)(p-c)} = S = \frac{1}{2} ab \sin C.$$ But $\sin C \leqslant 1,$ so $$2 = ab \sin C \leqslant ab.$$ Therefore $ab \geqslant 2.$ Equality hold for $x = 1, \, y = \sqrt 2 -1, \, z = 1.$

P/s. Using the AM-GM inequality, we have $$(x+y)(y+z) = xz+y(x+y+z) \geqslant 2\sqrt{xyz(x+y+z)} = 2.$$