10.1. (12 points) Find the minimum value of the function $f(x)=x^{2}+3 x+\frac{6}{x}+\frac{4}{x^{2}}-1$ on the ray $x>0$.
Answer: $3+6 \sqrt{2} \approx 11.49$.

Solution. For $x>0$ we have $t=x+\frac{2}{x} \geqslant 2 \sqrt{2}$ (equality is achieved at $x=\sqrt{2}$), and in this case $f(x)=\left(x+\frac{2}{x}\right)^{2}-3\left(x+\frac{2}{x}\right)-5$. Therefore, we need to find the minimum value of the function $g(t)=t^{2}-3 t-5$ for $t \geqslant 2 \sqrt{2}$. The function $g(t)$ is increasing for $t \geqslant \frac{3}{2}$, and since $2 \sqrt{2} \geqslant \frac{3}{2}$, the desired minimum value is $g(2 \sqrt{2})=3+6 \sqrt{2} \approx 11.49$.