Cho x,y,z là các số thực dương thỏa mãn x(x+y+z)=3yz.Cmr :(x+y)3+(x+z)3+3(x+y)(y+z)(z+x)≤5(y+z)3
Đặt y=ax;z=bx(a;b>0)⇒x2(1+a+b)=3abx2⇔a+b+1=3ab
Cần CM x3(a+1)3+x3(b+1)3+3x3(a+1)(b+1)(a+b)≤5x3(a+b)3
⇔(a+1)3+(b+1)3+3(a+1)(b+1)(a+b)≤5(a+b)3
⇔(a+b+2)3−3(a+b+2)(a+1)(b+1)+3(a+b)(a+1)(b+1)≤5(a+b)3
⇔(a+b+2)3−6(a+1)(b+1)≤5(a+b)3
⇔(a+b+2)3−2(3ab+3a+3b+3)≤5(a+b)3
⇔(a+b+2)3−2(4a+4b+4)≤5(a+b)3
Đặt t=a+b
BĐT ⇔(t+2)3−8(t+1)≤5t3⇔4t3−6t2−4≥0
⇔t(t−2)(2t+1)≥0
Ta có a+b+1=3ab≤34(a+b)2⇔3(a+b)2−4(a+b)−4≥0
⇔a+b≥2
Từ đó suy ra BĐT đúng