-- May I speak to Mr Smith?
-- , please. Here he comes.
（Ⅰ）设函数f（x）=|x﹣1a|+|x+a|（a＞0）．证明：f（x）≥2；
（Ⅱ）若实数x，y，z满足x2+4y2+z2=3，求证：|x+2y+z|≤3．
