Minimum value problem

Question

Find the minimum value of $(x+y)(y+z)$ where $x,y,z$ are positive real numbers satisfying the condition $$xyz(x+y+z)=1$$

Hint?

Answer 1

Just apply AM-GM.

$(x+y)(y+z)=xy+y^2+xz+yz=y(x+y+z)+zx=\frac y{xyz}+zx=zx+\frac1{zx}\geq 2$

Equality holds if and only if $zx=1$.

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Substituting $xz=1$, we have the condition that $y(x+y+\frac{1}{x}) = 1$. As such, $(x+y)(y+ \frac{1}{x}) = xy + \frac{y}{x} + y^2 + 1 = 1 + 1 = 2 $.

Answer 2

if you make the transformation to new variables $u,v,w$ with $$ u = y+z \\ v = z + x \\ w = x + y $$ then $$ u + v + w = 2(x+y+z) = 2s $$ and $$ x = s - u\\ y = s - v\\ z = s - w $$

so that $s$ is the semi-perimeter of a triangle with sides $u,v,w$ and the area of the triangle:

$$ A = \sqrt{s(s-u)(s-v)(s-w)} = 1 $$ and the problem is to minimize the product of two sides of a triangle with area of triangle constrained to be $1$

suppose $u$ is not smaller than either of the other two sides. then the minimum product of two sides in this case must be $vw$.

if, in $\mathbb{R}^2$ we let the base of the triangle be the interval $[-\frac{u}2,\frac{u}2]$ then to satisfy the constraint on area the third vertex must lie either of the lines $Y=\pm \frac2{u}$

a fairly straightforward argument shows that the minimum value of the product $vw$ occurs when the triangle is isosceles, hence (Pythagoras)

$$ v=w=\sqrt ( \left(\frac{u}2\right)^2 + \left(\frac2{u}\right)^2) $$ since then minimum of the expression inside the surd is the value 2, occurring when u=2, we have, as the sides of the triangle in minimal configuration, $2, \sqrt{2}, \sqrt{2}$, and the minimum value of the product required is $\sqrt{2} \times \sqrt{2} = 2$

Answer 3

Define the auxiliary function $F(x,y,z,\lambda) = (x+y)(y+z) + \lambda(1-xyz(x+y+z))$ and optimise it by taking the partial derivatives and settting them to zero.

But why can we do this?

This method applies Lagrange multipliers, since the level set, that is the constraint, is in the tangent space of the gradient. So we can set up this function, because we can make the claim that for $f(x,y,z) = (x+y)(y+z)$ and $g(x,y,z) = xyz(x+y+z)$, $\nabla f$ is parallel to $\nabla g$. Namely, $\exists \lambda$ such that $\nabla f + \lambda \nabla g = 0$.