10,11 |

Let $x, y, z$ be positive numbers and $x y z(x+y+z)=1$. Find the minimum value of the expression $(x+y)(x+z)$.
Make the substitutions $x=p-a, y=p-b, z=p-c$.

## Solution

Let $a=y+z, b=x+z, c=x+y, p=x+y+z$. Consider a triangle with sides $a, b, c$ (the triangle inequalities are obviously satisfied). The perimeter of this triangle is $2 p$, and let the area be denoted by $S$. By Heron's formula,

$S^{2}=p(p-a)(p-b)(p-c)=(x+y+z) x y z=1$, so $(x+y)(x+z)=b c \geq 2 S=2$. Equality is achieved for a right triangle

(for example, respectively,

## Answer

2.