Let a, b, and c be positive integers, and define three rational numbers x, y, and z as fol… — GRE General Test

Let

a

b

, and

c

be positive integers, and define three rational numbers

x

y

, and

z

as follows:

x = \frac{a}{a+b}, \quad y = \frac{b}{b+c}, \quad \text{and} \quad z = \frac{c}{c+a}

Which of the following statements MUST be true for all such positive integers

a

b

, and

c

? Select all that apply.

The sum $x + y + z$ satisfies the inequality $1 < x + y + z < 2$ .Cevap
The product $(1 - x)(1 - y)(1 - z)$ is identically equal to $xyz$ .Cevap
The product $xyz$ cannot exceed $\frac{1}{8}$ .Cevap
D
The sum $x + y + z$ can be equal to an integer.
E
If $x + y + z = \frac{3}{2}$ , then it must be that $x = y = z = \frac{1}{2}$ .

Cevap

The statements asserting that

1 < x + y + z < 2

, that

(1 - x)(1 - y)(1 - z) = xyz

, and that

xyz \le \frac{1}{8}

are all true.

The statement bounding the sum between 1 and 2 is correct because adjusting denominators to a common sum

a+b+c

reveals that the sum is strictly greater than 1 and strictly less than 2. The identity statement is correct because subtracting each fraction from 1 shifts the numerators cyclically without changing the overall product. The statement placing an upper bound of

\frac{1}{8}

on the product is correct by applying the AM-GM inequality to each denominator term.

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Establish strict bounds on the sum

x + y + z

Since

a, b, c > 0

, we have

\frac{a}{a+b} > \frac{a}{a+b+c}

. Summing all three terms gives

x + y + z > \frac{a+b+c}{a+b+c} = 1

. To find the upper bound, rewrite

\frac{a}{a+b} < \frac{a+c}{a+b+c}

. Summing all three terms gives

x + y + z < \frac{(a+c)+(a+b)+(b+c)}{a+b+c} = 2

. Thus,

1 < x + y + z < 2

This proves that the sum of the three fractions must always lie strictly between 1 and 2.

Compute the product

(1-x)(1-y)(1-z)

algebraically.

1 - x = 1 - \frac{a}{a+b} = \frac{b}{a+b}

1 - y = \frac{c}{b+c}

, and

1 - z = \frac{a}{c+a}

. The product is

\frac{b \cdot c \cdot a}{(a+b)(b+c)(c+a)} = \frac{abc}{(a+b)(b+c)(c+a)} = xyz

This proves the structural algebraic identity between

(1-x)(1-y)(1-z)

and

xyz

Apply the AM-GM inequality to find the upper bound of

xyz

Since

a+b \ge 2\sqrt{ab}

b+c \ge 2\sqrt{bc}

, and

c+a \ge 2\sqrt{ca}

, their product satisfies

(a+b)(b+c)(c+a) \ge 8\sqrt{a^2b^2c^2} = 8abc

. Taking the reciprocal gives

xyz = \frac{abc}{(a+b)(b+c)(c+a)} \le \frac{abc}{8abc} = \frac{1}{8}

This confirms that

xyz

has a maximum possible value of

\frac{1}{8}

(achieved when

a=b=c

Test counterexamples for the remaining statements.

Since

1 < x + y + z < 2

, no integer value is possible, ruling out integer sums. Additionally, substituting

a=1, b=2, c=1

gives

x=\frac{1}{3}, y=\frac{2}{3}, z=\frac{1}{2}

, whose sum is

\frac{3}{2}

despite

x, y, z

not being equal to

\frac{1}{2}

This disproves the statements claiming the sum can be an integer or that a sum of

\frac{3}{2}

requires all three fractions to be equal.

Anahtar Kavram

Bounding and algebraic properties of cyclic rational fractions

Tahmini Süre:3m 0s

Soru

Cevap

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