若不等式｜3x+2 ｜≥｜2x+a ｜对x∈R恒成立,则实数a的取值范围是?

|3x+2|≥|2x+a|
<==>9x^2+12x+4≥4x^2+4ax+a^2
<==>5x^2+4(3-a)x+(4-a^2)≥0
要使x∈R恒成立,即使判别式△≤0.
也即[4(3-a)]^2-20(4-a^2)≤0.
<==>9a^2-24a+16≤0
<==>(3a-4)^2≤0
<==>a=4/3.