Find out the answer to "Given two real numbers x, y so that..."

High School
Algebra
Algebra I
Quadratic function and equation

Janiyah Hays

Answered question

2022-03-27

Given two real numbers x, y so that

x^{2} + y^{2} + x y + 4 = 4 y + 3 x

. Prove that

3 (x^{3} - y^{3}) + 20 x^{2} + 2 x y + 5 y^{2} + 39 x \leq 100

Answer & Explanation

Ariel Cantrell

Beginner2022-03-28Added 8 answers

From the condition we obtain:
$y^{2} + (x - 4) y + x^{2} - 3 x + 4 = 0,$
which gives ${(x - 4)}^{2} - 4 (x^{2} - 3 x + 4) \geq 0$
or $0 \leq x \leq \frac{4}{3} .$
By the similar way we obtain:
$1 \leq y \leq \frac{7}{3},$
which gives:
$3 (x^{3} - y^{3}) + 20 x^{2} + 2 x y + 5 y^{2} + 39 x \leq$
$\leq \frac{64}{9} - 3 y^{3} + \frac{320}{9} + \frac{8}{3} y + 5 y^{2} + 52 .$
Thus, it's enough to prove that:
$\frac{64}{9} - 3 y^{3} + \frac{320}{9} + \frac{8}{3} y + 5 y^{2} + 52 \leq 100$
or $9 y^{3} - 15 y^{2} - 8 y + 16 \geq 0$
or $9 y^{3} - 24 y^{2} + 16 y + 9 y^{2} - 24 y + 16 \geq 0$
or ${(3 y - 4)}^{2} (y + 1) \geq 0$
The equality occurs for $x = y = \frac{4}{3},$ which says that 100 is a maximal such value.

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