例1 当x∈R时,关于x的不等式|x+7}≥m+2恒成立,求实数m的取值范围. 解因为函数y=|x+7|与Y=m+2的图象如图1所示,所以当m+2≤0时,符合题意,即m≤-2. 例2 当x∈R时,关于x的不等式|x-1|+|x+3|>a恒成立,求实数a的取值范围. Example 1 When x∈R, the inequality about x |x+7}≥m+2 holds true and the range of the real number m is obtained. The solution is due to the function y=|x+7| and Y=m+2. As shown in Figure 1, so when m + 2 ≤ 0, the meaning of the problem, that is, m ≤ -2. Case 2 When x ∈ R, the inequality about x | x-1 | + | x + 3 |> a is established, the value range of the real number a.