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: Rewrite each bracket using the constraint.
We are told \(x+y+z=1\), so \(1-x=y+z\), \(1-y=z+x\) and \(1-z=x+y\).
Since \(\dfrac{1}{x}-1=\dfrac{1-x}{x}\), we can replace \(1-x\) with \(y+z\), and do the same for the other two brackets.
So the expression becomes
\[ \left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)=\frac{(y+z)(z+x)(x+y)}{xyz} \]

Step 2: Apply AM-GM to each of the three sums.
The AM-GM rule says that for two positive numbers, their sum is at least twice the square root of their product, with equality only when the two numbers are equal.
Using this on each pair:
\[ y+z \geq 2\sqrt{yz}, \quad z+x \geq 2\sqrt{zx}, \quad x+y \geq 2\sqrt{xy} \]
Multiplying these three inequalities together:
\[ (y+z)(z+x)(x+y) \geq 8\sqrt{yz\cdot zx\cdot xy} = 8\sqrt{x^2y^2z^2} = 8xyz \]

Step 3: Divide both sides by \(xyz\).
\[ \frac{(y+z)(z+x)(x+y)}{xyz} \geq \frac{8xyz}{xyz} = 8 \]
So the whole expression can never fall below 8, for any positive \(x,y,z\) with \(x+y+z=1\).

Step 4: Check the value where equality happens.
Equality in AM-GM happens only when the two numbers in each pair are equal, so we need \(y=z\), \(z=x\) and \(x=y\) all at once, which forces \(x=y=z\).
Since \(x+y+z=1\), this gives \(x=y=z=\dfrac{1}{3}\), a valid choice of positive real numbers.
At this point, \(\dfrac{1}{x}-1=3-1=2\) for each bracket, so the product is \(2\times2\times2=8\).
This confirms the bound of 8 is actually reached, not just a theoretical limit.

Step 5: Rule out the other options.
4 and 16 are not correct: we proved the expression is always at least 8, so 4 is too small to ever be the least value, and 16 is bigger than a value (8) that is genuinely reached, so 16 cannot be the least value either.
Since 8 is both a valid lower bound and an achievable value, it is exactly the least value, so "None of the above" is wrong too.

Final Answer:
The least value of the expression is 8, reached at \(x=y=z=\tfrac{1}{3}\).
\[ \boxed{8} \]