Question

已知x＞0，y＞0，求证x*2/y+y*2/x≥x+y

Answer 1

要证x*2/y+y*2/x≥x+y，即x*2/y+y*2/x-（x+y)≥0
通分得(x^3+y^3-x^2y-y^2x)/xy≥0
化简，(x^2(x-y)-y^2(x-y))/xy≥0
(x+y)(x-y)(x-y)/xy≥0
因为x＞0，y＞0，所以(x+y)(x-y)(x-y)/xy≥0
原题得证

Answer 2

证：
x²/y+y²/x-(x+y)
=[x²(x-y)-y²(x-y)]/(xy)
=(x-y)(x²-y²)/(xy)
=(x-y)²(x+y)/(xy)
由x>0 y>0得x+y>0 xy>0
平方项恒非负，(x-y)²≥0
(x-y)²(x+y)/(xy)≥0
x²/y+y²/x-(x+y)≥0
x²/y+y²/x≥x+y 当x=y时取等号。