Cho $x, y, z$ thoả mãn $x^2+y^2+z^2=2$
Tìm min, max của $P=x^3+y^3+z^3-3xyz$
đề hay đó –  cuungonghinh 09-11-12 10:28 PM
dạng này mình bó tay luôn –  babylionneu 09-11-12 10:13 PM
bài này quen quen –  banhquykeomut 09-11-12 10:00 PM
Ta có:
     $(x^3+y^3+z^3-3xyz)^2$
$=(x(x^2-yz)+y(y^2-xz)+z(z^2-xy))^2$
$\le(x^2+y^2+z^2)((x^2-yz)^2+(y^2-xz)^2+(z^2-xy)^2)$
$=(x^2+y^2+z^2)(x^4+y^4+z^4+x^2y^2+y^2z^2+z^2x^2-2xyz(x+y+z))$
$=(x^2+y^2+z^2)((x^2+y^2+z^2)^2-(xy+yz+zx)^2)$
$\le(x^2+y^2+z^2)^3=8$
Suy ra: $-2\sqrt2\le x^3+y^3+z^3-3xyz\le 2\sqrt2$
Min$P=-2\sqrt2$ chẳng hạn khi: $(x,y,z)=(-\sqrt2,0,0)$
Max$P=2\sqrt2$ chẳng hạn khi: $(x,y,z)=(\sqrt2,0,0)$

cảm ơn Ad nhé –  daovandungtphcm 10-11-12 06:51 PM
$\textbf{Cách 2}$
Đặt
$A=(x+y+z)^2, B=x^2+y^2+z^2-xy-yz-zx$.
Ta có $A+2B=3(x^2+y^2+z^2)=6$.
Ta có
$P^2=AB^2=A.B.B \underbrace{\le}_{Cô-si} \left (\frac{A+B+B}{3} \right )^3= \left (\frac{6}{3} \right )^3=8$
Suy ra $-2\sqrt 2 \le P \le 2\sqrt 2$.
Vậy
Min$P=-2\sqrt2\Leftrightarrow A=B\Leftrightarrow xy+yz+zx=0$ chẳng hạn khi: $(x,y,z)=(-\sqrt2,0,0)$
Max$P=2\sqrt2\Leftrightarrow A=B\Leftrightarrow xy+yz+zx=0$ chẳng hạn khi: $(x,y,z)=(\sqrt2,0,0)$

Hãy ấn chữ V dưới đáp án để chấp nhận nếu như bạn thấy lời giải này chính xác, và nút mũi tên màu xanh để vote up nhé. Thanks! –  Trần Nhật Tân 09-11-12 04:08 PM

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