Given two real numbers x, y so that

Janiyah Hays

Janiyah Hays

Answered question

2022-03-27

Given two real numbers x, y so that x2+y2+xy+4=4y+3x. Prove that 3(x3y3)+20x2+2xy+5y2+39x100.

Answer & Explanation

Ariel Cantrell

Ariel Cantrell

Beginner2022-03-28Added 8 answers

From the condition we obtain:
y2+(x4)y+x23x+4=0,
which gives (x4)24(x23x+4)0
or 0x43.
By the similar way we obtain:
1y73,
which gives:
3(x3y3)+20x2+2xy+5y2+39x
6493y3+3209+83y+5y2+52.
Thus, it's enough to prove that:
6493y3+3209+83y+5y2+52100
or 9y315y28y+160
or 9y324y2+16y+9y224y+160
or (3y4)2(y+1)0
The equality occurs for x=y=43, which says that 100 is a maximal such value.

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