Question

已知x^3+y^3=2，求证:x+y〈=2

数学题

Answer 1

(x+y)(x2+y2-xy)=2
(x+y)[(x+y)^2-3xy]=2
x+y≥2√xy
xy≤1/4*(x+y)^2
-3xy≥-3/4*(x+y)^2
(x+y)[(x+y)^2-3xy]≥1/4*(x+y)^3
2≥1/4*(x+y)^3
1/4*(x+y)^3≤2
(x+y)^3≤8
(x+y)≤2

Answer 2

x^3+y^3=2 (x+y)(x2-xy+y2）=2 （x+y)((x+y)2-3xy]=2
又 (x+y)2≥4xy, 所以 (x+y)2-3xy≥ (x+y)2/4
所以 x+y≤0，或2=（x+y)((x+y)2-3xy]≥ (x+y)3/4 0< x+y〈=2
x+y〈=2