Seeking the minimum without calculus

Question

Imagine booking a ticket to the first octant of $\,\mathbb R^3$ and landing on the surface given by $$xyz\:=\:x+y+z+2\,.\tag{S}$$ This defining constraint for $\,x,y,z\geqslant0\,$ is equivalent to $$\frac 1{x+1}+\frac1{y+1}+\frac1{z+1}\:=\:1$$ which reminds a bit of conjugate exponents in two dimensions (that is $\frac 1p+\frac 1q=1$).

My pair of questions (the second of which depends on the first one):

• Does either side of (S), say $h:\mathbb R^3\to\mathbb R, (x,y,z)\mapsto xyz$, when restricted to the surface, has a unique absolute minimum in $(2,2,2)$?
• May this be proved along an "algebraic track", that is without using Lagrange multipliers?

I came across (S) during the analysis of this inequality question on this SE site.

Answer 1

By the AM-HM inequality:

$$ 3 = \frac{3}{\cfrac 1{x+1}+\cfrac1{y+1}+\cfrac1{z+1}} \le \frac{x+1+y+1+z+1}{3} = 1 + \frac{x+y+z}{3} = 1 + \frac{xyz - 2}{3} $$

Therefore $\,xyz \ge 8\,$ with equality iff $\,x=y=z=2\,$.

Answer 2

We can use the AM-GM inequality instead of Lagrange multipliers. Let $a^3 = xyz$. Then we have that

$$ a^3 = xyz = x + y + z + 2 \geq 3 \sqrt[3]{xyz} + 2 = 3a + 2. $$ Equality occurs if and only if $x = y = z$.

We thus have that $$ a^3 - 3a - 2 \geq 0 $$ which is equivalent to $$ (a + 1)^2 (a - 2) \geq 0 $$ and so $a \geq 2$.

We thus have that $xyz \geq 2^3 = 8$, and for equality to occur, we must have that $x = y = z = 2$.