Use the Constraint!

Algebra Level pending

x 3 y + y x 3 − 2 = ∣ 2 − x − y ∣ \large x^3y+y x^3-2=| \sqrt{2-x-y} |

Find all real ( x , y ) (x,y) solutions to the equation above. If your solution set is ( x 1 , y 1 ) (x_1,y_1) ; ( x 2 , y 2 ) (x_2,y_2) ; . . . ( x k , y k ) ...(x_k,y_k) then submit your answer as ∑ i k ( x i + y i ) \displaystyle \sum_{i}^{k} \left(x_i+y_i \right) .


The answer is 2.00.

Kushal Bose by Kushal Bose , 28, India

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1 solution

Kushal Bose
May 1, 2017

If the system has real solutions then 2 − x − y ≥ 0 ⟹ x + y ≤ 2 2-x-y \geq 0 \implies x+y \leq 2

Using A.M.-G.M. x + y ≥ 2 x y x+y \geq 2 \sqrt{x y} .So, 2 ≥ x + y ≥ 2 x y ⟹ x y ≤ 1 2 \geq x+y \geq 2 \sqrt{xy} \implies xy \leq 1

Now consider the expression x 3 y + y 3 x = x y ( x 2 + y 2 ) = x y [ ( x + y ) 2 − 2 x y ] = x y ( x + y ) 2 − 2 ( x y ) 2 x^3 y+y^3 x=xy(x^2+y^2) \\ = xy[(x+y)^2-2 xy] \\ =xy (x+y)^2-2 (xy)^2

Using the two inequalities x + y ≤ 2 x+y \leq 2 and x y ≤ 1 xy \leq 1

x y ( x + y ) 2 ≤ 1 × 4 = 4 xy (x+y)^2 \leq 1 \times 4=4 and 2 ( x y ) 2 ≤ 2 2 (xy)^2 \leq 2

So, x y ( x + y ) 2 − 2 ( x y ) 2 ≤ 2 ⟹ x 3 y + y 3 x ≤ 2 ⟹ x 3 y + y 3 x − 2 ≤ 0 xy (x+y)^2-2 (xy)^2 \leq 2 \implies x^3 y+y^3 x \leq 2 \implies x^3 y+y^3 x-2 \leq 0

This says ∣ 2 − x − y ∣ ≤ 0 | \sqrt{2-x-y}| \leq 0 which is only possible if the equality holds (which is equal to zero).

So,the only solution is x = y = 1 x=y=1

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