Dùng BĐT $AM-GM$ chứng minh rằng: $$3\left(x^2+y^2+z^2\right)\geq\left(x+y+z\right)^2$$
bài này ko khó lắm –  babylionneu 07-01-13 08:49 PM
Vì $x,y,z$ không có điều kiện dương nên ta cần lưu ý một chút. Cụ thể như sau
Áp dụng AM-GM cho $x^2,y^2 \ge 0$ ta được
$x^2+y^2 \ge 2\sqrt{x^2y^2}=2|xy| \ge 2xy$
Tương tự
$z^2+y^2\ge 2zy$
$x^2+z^2\ge 2xz$
 Cộng theo từng vế ba BĐT trên ta có
$2(x^2+y^2+z^2) \ge 2xy+2yz+2zx$
từ đây suy ra đpcm.
Đẳng thức xảy ra $\Leftrightarrow \begin{cases}x=y=z\\ x,y,z \text{ cùng dấu}  \end{cases}\Leftrightarrow x=y=z$
lời giải hay nhỉ, hì –  babylionneu 07-01-13 08:49 PM
Theo AM-GM thì
$2xy\leq x^2+y^2$
$2zy\leq z^2+y^2$
$2xz\leq x^2+z^2$
Vậy $x^2+y^2+z^2+2xy+2yz+2zx\leq 3(x^2+y^2+z^2)$ 
$\Leftrightarrow (x+y+z)^2\leq 3(x^2+y^2+z^2)$ 

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