Help with algebra olympiad question

[hollow moth]

Prove that for any natural numbers x, y (x ≥ 3, y ≥ 3) the following inequality is true: x²y² − 144xy + 432(x+y) ≥ 1296

[dense locust]

$-144xy+432x+432y-1296 \ge -x^2y^2$

[real stratus]

Schlaumau

[dense locust]

$144xy-36\cdot 12x-36\cdot 12y+36^2 \leq x^2y^2$

[real stratus]

Schlaumau

[dense locust]

$(12x-36)(12y-36) \leq x^2y^2$

[real stratus]

Schlaumau

[dense locust]

Now we have $(x-6)^2 \geq 0$

[real stratus]

Schlaumau

[dense locust]

$x^2 \geq 12x-36$

same thing goes for y

multiply these inequalities and you get the result

@hollow moth

[real stratus]

Schlaumau

[dense locust]

multiplication works because 12x-36 and 12y-36 are always greater or equal to 0

[hollow moth]

Thank you very much, though I don't understand how you did this and why?

[manic lynx]

@hollow moth has given 1 rep to @dense locust

[hollow moth]

I understood the rest

[dense locust]

I factored the left side of the inequality. I you expand (12x-36)(12y-36) you will get 144xy - 36*12x - 36*12y + 36^2 again. In some cases you will just have to guess the factors and then see if it works like it does in this case. I did this because it allowed me to use x^2>=12x-36 to show that the inequality holds.