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Ta có:
⇔u3−3u−2≤v3−3v+2⇔ (u+1)2(u−2)≤(v−1)2(v+2)
Đặt x=u+1,y=v−1 thì:
BĐT ⇔ x3−3x2≤y3+3y2⇔ x3−y3≤3(x2+y2)
Ta có: x−y=(u−v)+2≤2
⇒(x−y)(x2+xy+y2)≤2(x2+xy+y2)=2(x2+y2)+2xy
≤2(x2+y2)+(x2+y2)=3(x2+y2)⇒x3−y3≤3(x2+y2)( đpcm)
Dấu bằng xảy ra ⇔ x=y=0⇔ u=−1,v=1
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