Inequalities?

Algebra Level 3

Positive reals x x and y y are such that

101 x 3 − 2019 x y + 101 y 3 = 100 101x^3-2019xy+101y^3=100

Find the maximum integer value of x y xy .


The answer is 100.

Mohammed Imran by Mohammed Imran , 13, India

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1 solution

x y = 101 ( x 3 + y 3 ) − 100 2019 xy=\dfrac {101(x^3+y^3)-100}{2019}

≥ 202 ( x y ) 3 2 − 100 2019 \geq \dfrac {202(xy)^{\frac{3}{2}}-100}{2019}

⟹ 40804 ( x y ) 3 − 4076361 ( x y ) 2 − 403800 ( x y ) − 10000 ≤ 0 \implies 40804(xy)^3-4076361(xy)^2-403800(xy)-10000\leq 0

⟹ x y ≤ 100 \implies xy\leq 100 .

So the maximum of x y xy is 100 \boxed {100} .

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In questions like these, you also have to show the equality case. Here it works for x = y = 10 x=y=10 .

Vilakshan Gupta - 11 months, 1 week ago

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