Multiply both sides of the equation by $6$, the least common multiple of $3,2$. Since $6$ is positive, the inequality direction remains the same.
$$2\left(3x+1\right)<3x\left(1+x\right)$$
Use the distributive property to multiply $2$ by $3x+1$.
$$6x+2<3x\left(1+x\right)$$
Use the distributive property to multiply $3x$ by $1+x$.
$$6x+2<3x+3x^{2}$$
Subtract $3x$ from both sides.
$$6x+2-3x<3x^{2}$$
Combine $6x$ and $-3x$ to get $3x$.
$$3x+2<3x^{2}$$
Subtract $3x^{2}$ from both sides.
$$3x+2-3x^{2}<0$$
Multiply the inequality by -1 to make the coefficient of the highest power in $3x+2-3x^{2}$ positive. Since $-1$ is negative, the inequality direction is changed.
$$-3x-2+3x^{2}>0$$
To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$-3x-2+3x^{2}=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $3$ for $a$, $-3$ for $b$, and $-2$ for $c$ in the quadratic formula.
For the product to be positive, $x-\left(\frac{\sqrt{33}}{6}+\frac{1}{2}\right)$ and $x-\left(-\frac{\sqrt{33}}{6}+\frac{1}{2}\right)$ have to be both negative or both positive. Consider the case when $x-\left(\frac{\sqrt{33}}{6}+\frac{1}{2}\right)$ and $x-\left(-\frac{\sqrt{33}}{6}+\frac{1}{2}\right)$ are both negative.
