Inequalities (Olympiad)mediumPYQ 2015 · RMOFree

Inequalities (Olympiad): Real Numbers (RMO 2015)

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

Question

The all real numbers

x, y

such that

x^2 + 2y^2 + \dfrac{1}{2} \le x(2y + 1)

(x, y) = \left(4,\ \dfrac{17}{2}\right)

(x, y) = \left(3,\ \dfrac{1}{2}\right)

(x, y) = \left(1,\ \dfrac{1}{2}\right)

.correct

(x, y) = \left(-1,\ \dfrac{-1}{2}\right)

Solution

Step 1: Multiply by

2

and move to one side.

2x^2 + 4y^2 + 1 \le 2x(2y + 1) \implies 2x^2 + 4y^2 + 1 - 4xy - 2x \le 0.

Step 2: Regroup into squares.

(x^2 - 4xy + 4y^2) + (x^2 - 2x + 1) \le 0 \implies (x - 2y)^2 + (x - 1)^2 \le 0.

Step 3: A sum of squares is at most zero only if each is zero:

x - 2y = 0

and

x - 1 = 0

. Step 4: Hence

x = 1

and

y = \dfrac{x}{2} = \dfrac{1}{2}

. Answer:

(x, y) = \left(1,\ \dfrac{1}{2}\right)

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