Example 2 $x$ is a positive real number, find the minimum value of the function $y=x^{2}+x+\frac{3}{x}$.
By substituting some simple values of $x$ such as $\frac{1}{2}, 1, 3$, one can find that the given function is not monotonic, which often makes it difficult to approach at first glance. However, with some problem-solving experience, one would realize that such problems can be easily solved by appropriately transforming the function.

Solution 1: $y=x^{2}-2 x+1+3 x-6+\frac{3}{x}+5=(x-1)^{2}+3\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} +5 \geqslant 0+0+5=5$, and there exists $x_{0}=1$ such that $y=5$, therefore the minimum value of $y$ is 5.

Solution 2: $y=x^{2}+x+\frac{1}{x}+\frac{1}{x}+\frac{1}{x} \geqslant 5\left(x^{2} \cdot x \cdot \frac{1}{x} \cdot \frac{1}{x} \cdot \frac{1}{x}\right)^{\frac{1}{5}}$ (Arithmetic-Geometric Mean Inequality $)=5$, and there exists $x_{0}=1$ such that $y=5$, therefore the minimum value of $y$ is 5.