Prove that $(x-1)(y-1)(z-1)\geq 8$.

Question

Let $x, y, z$ be positive integers, such that $\frac{1}{x}+ \frac{1}{y}+ \frac{1}{z} \leq 1$. Find $\inf_{x, y, z} (x-1)(y-1)(z-1)$.

After a few trials, I'd say the answer is $8$, but I can't prove it. Using Am-GM I wasn't even able to prove the $\inf$ was $\geq 8$.

Answer 1

Expand $(x-1)(y-1)(z-1)$. We get $$xyz-(xy+yz+xz)+(x+y+z)-1.\tag{1}$$

By the constraint $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le 1$, the sum of the first two terms in (1) is $\ge 0$.

So we want to show that $x+y+z\ge 9$. This follows from the Arithmetic Mean Harmonic Mean Inequality. We have equality only at $x=y=z=3$.

Answer 2

Let $\frac{1}{x}=a, \frac{1}{y}=b, \frac{1}{z}=c$. Then $a+b+c\le 1$. WOLOG, let us suppose $a+b+c=1$. We need to show that

$$ \Pi \frac{a+b}{c} \ge 8 $$

Now you can use Am-GM.

Answer 3

the product is monotonically increasing in each variable. so can't we consider triples starting with $(1,1,1)$ eliminating families of triples violating the reciprocal sum on the way?

wlog $x \leq y \leq z$.

this way we have $x,y,z > 1$, $x = 2 \rightarrow y \geq 3, z \geq 4$, $x = 2 \land y = 3 \rightarrow z \geq 6$, $x = 2 \land y = 4 \rightarrow z \geq 4$, and therefore minimal elements $\{(2,3,6),(2,4,4),(3,3,3)\}$ for the set of valid triples $(x, y, z)$ and the partial order $(x_0,y_0,z_0) \leq_{t} (x_1,y_1,z_1) \equiv x_0 \leq x_1 \land y_0 \leq y_1 \land z_0 \leq z_1 $.

$(2-1)(3-1)(6-1) = 10, (2-1)(4-1)(4-1) = 9, (3-1)(3-1)(3-1) = 8$ which completes the proof.