Решить неравенство √(x+3)< √(x-1)+√(x-2)

ОДЗ
x+3≥0⇒x≥-3
x-1≥0⇒x≥1
x-2≥0⇒x≥2
x∈[2;∞)
x+3<x-1+2√(x²-3x+2)+x-2
x+3-2x+3<2√(x²-3x+2)
6-x<2√(x²-3x+2)
36-12x+x²<4x²-12x+8
4x²-12x+8-36+12x-x²>0
3x²-28>0
(x-2√21/3)(x+2√21/3)>0
x<-2√21/3 U x>2√21/3
x∈(2√21/3;∞)