gt⇔5x2+5(y2+z2)−9x(y+z)−18yz=0(1)Ta có:yz≤14(y+z)2;y2+z2≥12(y+z)2
⇒18yz−5(y2+z2)≤2(y+z)2(2)
Từ(1)&(2)⇒5x2−9x(y+z)≤2(y+z)2
⇔(x−2(y+z))(5x+y+z)≤0
⇒x≤2(y+z)
P≤2x(y+z)2−1(x+y+z)3≤4y+z−127(y+z)3
Đặt t=1y+z⇒P≤4t-t327
Ta phải cm P≤16
⇔(t+12)(t−6)227≥0(luôn đúng)
Dấu''='' xra⇔x=13;y=z=112