Cho $x, y, z$ không âm thỏa mãn: $x^2+y^2+z^2=3$. Tìm max: $P=xy+yz+zx+\frac{4}{x+y+z}$
Đặt $a=x+y+z,b=xy+yz+zx$
Ta có $a^2=3+2(xy+yz+zx) \ge 3\Leftrightarrow a \ge \sqrt 3$
Mặt khác $a=x+y+z \le \sqrt{3(x^2+y^2+z^2)}=3$
$\Rightarrow a \in[\sqrt3;3]$
Lại có  $x^2+y^2+z^2=a^2-2b\Leftrightarrow a^2-2b=3\Leftrightarrow b=\frac{a^2-3}{2}$
~~~~~~
$P=b+\frac 4a=\frac{a^2-3}{2}+\frac 4a=\frac{a^3+8}{2a}-\frac 32$
Xét $f(a)=\frac{a^3+8}{2a}$ trên $[\sqrt3; 3]$ tìm đc max $= \frac{35}{6}$ khi $a=3$
$P \le\frac{13}3$
Khi thay $x=y=z=1$ thì $P= \frac{13}3$
Vậy $\max P=\frac{13}3$ khi $x=y=z=1$
ý anh là, dấu bằng phải xẩy ra ở mọi chỗ –  ๖ۣۜDevilღ 12-05-16 10:37 PM
$x=y=1 $=> a=3 có liên quan gì nó đâu a –  tran85295 12-05-16 10:36 PM
nhưng dấu bằng lại x=y=z=1 –  ๖ۣۜDevilღ 12-05-16 10:35 PM
à chỉ cần x=0,y=0 là đủ rồi :)) –  tran85295 12-05-16 10:34 PM
thế tý dấu bằng xẩy ra thì cái xy yz zx phải bằng 0 chứ nhỉ –  ๖ۣۜDevilღ 12-05-16 10:33 PM
e thấy bài trên kia cũng đánh giá y như vậy :3 –  tran85295 12-05-16 10:30 PM
em xem lại thử –  ๖ۣۜDevilღ 12-05-16 10:29 PM
sai sao ạ –  tran85295 12-05-16 10:27 PM
đoạn đánh giá đầu tiên bị Sai –  ๖ۣۜDevilღ 12-05-16 10:26 PM

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