Решите неравенство Пожалуйста помогите.....

Решите задачу:

$\sqrt{3x+1}\leq x+1\; \; \; \Leftrightarrow \; \; \; \left\{\begin{array}{l}x+1\geq 0\\3x+1\geq 0\\3x+1\leq (x+1)^2\end{array}\right \\\\3x+1\leq x^2+2x+1\\\\x^2-x\geq 0\\\\x(x-1)\geq 0\; \; ,\; \; \; +++[\, 0\, ]---[\, 1\, ]+++\\\\x\in (-\infty ,\, 0\, ]\cup [\, 1,+\infty )\\\\\\\left\{\begin{array}{l}x\geq -1\\x\geq -\frac{1}{3}\\x\in (-\infty ,\, 0\, ]\cup [\, 1,+\infty )\end{array}\right\; \; \; \Rightarrow \; \; \; \underline {\; x\in [-\frac{1}{3}\, ,\, 0\, ]\cup [\, 1,+\infty )\; }$