Question

If \(x + y + z = 1\) and \(x, y, z\) are positive real numbers, then the least value of \(\left(\dfrac{1}{x}-1\right)\left(\dfrac{1}{y}-1\right)\left(\dfrac{1}{z}-1\right)\) is:

• A. 4
• B. 8
• C. 16
• D. None of the above

Hint:

Rewrite each \(\frac{1}{x}-1\) as \(\frac{y+z}{x}\) using \(x+y+z=1\), then apply AM-GM to \((y+z)(z+x)(x+y)\) to find the least value.

Answer 1

The Correct Option is B

Step 1: Set up the problem for optimization.
We want the smallest value of $f(x,y,z)=\left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)$ on positive numbers with $x+y+z=1$.
A good way to check where a smooth function like this is smallest is to first look for a point where all the partial slopes balance out (the "critical point"), and then check the behavior near the edges of the region.

Step 2: Look for a symmetric critical point.
Because the expression treats $x$, $y$ and $z$ in exactly the same way, if there is a single interior point where the function is smallest, it is very likely the symmetric point $x=y=z$.
With $x+y+z=1$, symmetry gives $x=y=z=\frac{1}{3}$.
At this point each bracket is $\frac{1}{1/3}-1=3-1=2$, so $f=2\times2\times2=8$.

Step 3: Check what happens near the edges.
Since $x,y,z$ are positive and add to 1, none of them can be 0, but any one of them, say $z$, can get arbitrarily close to 0 while $x+y$ stays close to 1.
As $z\to 0^{+}$, the bracket $\frac{1}{z}-1\to\infty$, while the other two brackets stay finite (bounded, since $x$ and $y$ stay bounded away from 0).
So the product $f(x,y,z)\to\infty$ as any variable shrinks toward 0.

Step 4: Conclude which point gives the minimum.
Since $f$ blows up to infinity near every edge of the region, and $f$ is smooth inside, the value at the one balanced interior point we found, $x=y=z=\frac{1}{3}$, must be the smallest value $f$ ever takes.
That value is 8.

Step 5: Match with the options.
8 is exactly option (B), so the answer is 8, not 4, not 16, and "None of the above" does not apply because 8 is genuinely on the list.

Final Answer:
The least value is 8.
$\boxed{8}$