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>1x > 1, y>1y > 1, z>1z > 1 satisfying
x+y+z+3x−1+3y−1+3z−1=2(x+2+y+2+z+2).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).
Ax=y=z=3+132x = y = z = \dfrac{3 + \sqrt{13}}{2}.correct
Bx=y=z=2+132x = y = z = \dfrac{2+ \sqrt{13}}{2}.
Cx=y=z=4+132x = y = z = \dfrac{4 + \sqrt{13}}{2}.
Dx=y=z=5+132x = y = z = \dfrac{5 + \sqrt{13}}{2}.
Solution
Step 1: Move everything to one side and group by variable:
∑x, y, z(x+3x−1−2x+2)=0.\sum_{x,\,y,\,z} \left(x + \frac{3}{x - 1} - 2\sqrt{x + 2}\right) = 0.
 Step 2: Combine each bracket over x−1x - 1:
x+3x−1−2x+2=x2−x+3−2(x−1)x+2x−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 x2−x+3=(x−1)2+(x+2)x^2 - x + 3 = (x - 1)^2 + (x + 2), the numerator is
(x−1)2+(x+2)−2(x−1)x+2=[(x−1)−x+2]2.(x - 1)^2 + (x + 2) - 2(x - 1)\sqrt{x + 2} = \big[(x - 1) - \sqrt{x + 2}\big]^2.
 Hence
x+3x−1−2x+2=[(x−1)−x+2]2x−1.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>1x, y, z > 1, each denominator is positive, so each term is non-negative. Their sum is zero only if each numerator vanishes:
(x−1)−x+2=0for each of x,y,z.(x - 1) - \sqrt{x + 2} = 0 \quad \text{for each of } x, y, z.
 Step 5: Solve t−1=t+2t - 1 = \sqrt{t + 2} for t>1t > 1. Squaring, t2−2t+1=t+2t^2 - 2t + 1 = t + 2, so t2−3t−1=0t^2 - 3t - 1 = 0, giving t=3±132t = \dfrac{3 \pm \sqrt{13}}{2}. Since t>1t > 1, only t=3+132t = \dfrac{3 + \sqrt{13}}{2} is valid. Answer: x=y=z=3+132x = y = z = \dfrac{3 + \sqrt{13}}{2}.
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