16.1. [9.4 (15 points), 10.4 (15 points)] On the graph of the function $y=x+\frac{1}{x}$, where $x>0$, find the point closest to the origin.
Answer: $\left(\frac{1}{\sqrt[4]{2}} ; \frac{1+\sqrt{2}}{\sqrt[4]{2}}\right)$

Solution. The square of the distance from the point $\left(x, x+\frac{1}{x}\right)$ to the origin is

$$
r^{2}(x)=x^{2}+\left(x+\frac{1}{x}\right)^{2}=2 x^{2}+2+\frac{1}{x^{2}}
$$

By the inequality between the arithmetic mean and the geometric mean, we get

$$
r^{2}(x)=2 x^{2}+\frac{1}{x^{2}}+2 \geqslant 2 \sqrt{2 x^{2} \cdot \frac{1}{x^{2}}}+2=2 \sqrt{2}+2
$$

with equality achieved when $2 x^{2}=\frac{1}{x^{2}}$, i.e., $x^{4}=\frac{1}{2}$. Since $x>0$, from this we find $x=\frac{1}{\sqrt[4]{2}}, y=x+\frac{1}{x}=\frac{1+\sqrt{2}}{\sqrt[4]{2}}$.