Algebra (Olympiad)hardFree

Algebra (Olympiad): Determine Real Numbers Satisfying

JEE Maths question with a full step-by-step solution.

Question

Determine all real numbers

x > 1

y > 1

z > 1

satisfying

x + y + z + \frac{3}{x - 1} + \frac{3}{y - 1} + \frac{3}{z - 1} = 2\big(\sqrt{x + 2} + \sqrt{y + 2} + \sqrt{z + 2}\big).

x = y = z = \dfrac{3 + \sqrt{13}}{2}

.correct

x = y = z = \dfrac{2+ \sqrt{13}}{2}

x = y = z = \dfrac{4 + \sqrt{13}}{2}

x = y = z = \dfrac{5 + \sqrt{13}}{2}

Solution

Step 1: Move everything to one side and group by variable:

\sum_{x,\,y,\,z} \left(x + \frac{3}{x - 1} - 2\sqrt{x + 2}\right) = 0.

Step 2: Combine each bracket over

x - 1

x + \frac{3}{x - 1} - 2\sqrt{x + 2} = \frac{x^2 - x + 3 - 2(x - 1)\sqrt{x + 2}}{x - 1}.

Step 3: Since

x^2 - x + 3 = (x - 1)^2 + (x + 2)

, the numerator is

(x - 1)^2 + (x + 2) - 2(x - 1)\sqrt{x + 2} = \big[(x - 1) - \sqrt{x + 2}\big]^2.

Hence

x + \frac{3}{x - 1} - 2\sqrt{x + 2} = \frac{\big[(x - 1) - \sqrt{x + 2}\big]^2}{x - 1}.

Step 4: As

x, y, z > 1

, each denominator is positive, so each term is non-negative. Their sum is zero only if each numerator vanishes:

(x - 1) - \sqrt{x + 2} = 0 \quad \text{for each of } x, y, z.

Step 5: Solve

t - 1 = \sqrt{t + 2}

for

t > 1

. Squaring,

t^2 - 2t + 1 = t + 2

, so

t^2 - 3t - 1 = 0

, giving

t = \dfrac{3 \pm \sqrt{13}}{2}

. Since

t > 1

, only

t = \dfrac{3 + \sqrt{13}}{2}

is valid. Answer:

x = y = z = \dfrac{3 + \sqrt{13}}{2}

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