câu 9gt
⇔5x2+5(y2+z2)−9x(y+z)−18yz=0 yz≤14(y+z)2;y2+z2≥12(y+z)2
⇒18yz−5(y2+z2)≤2(y+z)2⇒5x2−9x(y+z)≤2(y+z)2
⇔(x−2(y+z))(5x+y+z)≤0
⇒x≤2(y+z)
P=xy2+z2−1(x+y+z)3≤2x(y+z)2−1(x+y+z)3
≤4y+z−127(y+z)3 Đặt 1y+z=t
⇒P≤4t−127t3
Ta sẽ chứng minh 4t−127t3≤16
⇔(t−6)2(t+12)27≥0 (đúng ∀t>0)
⇒P≤16
dấu '=" ⇔x=13;y=z=112