For what real number $x$ does the following expression attain its minimum?

$$
\sqrt{x^{2}+(x+1)^{2}}+\sqrt{x^{2}+(x-1)^{2}}
$$
I. solution. Let us denote the expression under investigation by $f(x)$. The function $f(x)$ is defined for all real numbers $x$, since the square roots contain the sum of squares of different numbers, which is always positive. Thus, the value of $f(x)$ is always positive. Squaring is a monotonic operation among non-negative numbers, so $f^{2}(x)$ is minimal where $f(x)$ is.

$$
f^{2}(x)=\left(\sqrt{2 x^{2}+1+2 x}+\sqrt{2 x^{2}+1-2 x}\right)^{2}=4 x^{2}+2+2 \sqrt{4 x^{4}+1}
$$

It can be directly determined from this expression that its smallest value is taken at $x=0$.

Piros Sándor (Debrecen, Fazekas M. Gym., III. grade)

II. solution. Consider the triangle whose vertices are $(x, y)$ in the Cartesian coordinate system of the plane $A(0 ;-1), B(x ; x), C(0 ; 1)$ (Fig. 1). By the triangle inequality,

$$
A C \leq A B+B C
$$

Expressing the distances in the inequality using coordinates,

$$
2 \leq \sqrt{x^{2}+(x+1)^{2}}+\sqrt{x^{2}+(x-1)^{2}}
$$

follows. Equality holds if and only if the triangle $ABC$ is degenerate, and $B$ is a point on the segment $AC$.

![](https://cdn.mathpix.com/cropped/2024_05_02_2a6bc5d90a9d0ff5e218g-1.jpg?height=469&width=380&top_left_y=955&top_left_x=859)

As $x$ varies, $B$ moves along the line $y=x$. Therefore, equality holds if and only if $B$ is at the origin, i.e., $x=0$. Since the expression on the right side of the inequality is precisely the expression under investigation, this answers the question of the problem.

The minimum value on the left side of the inequality also emerges almost by itself.

Ármós Lajos (Debrecen, Fazekas M. Gym., IV. grade)

Remark. Based on the provided solution, the problem can be easily generalized.

III. solution (outline). We solve the problem using differential calculus. The function $f(x)=\sqrt{x^{2}+(x+1)^{2}}+\sqrt{x^{2}+(x-1)^{2}}$ is defined and continuous for all real numbers $x$, and its derivative is

$$
f^{\prime}(x)=\frac{2 x+1}{\sqrt{2 x^{2}+2 x+1}}+\frac{2 x-1}{\sqrt{2 x^{2}-2 x+1}}
$$

This is also defined and continuous everywhere.

Solving the equation $f^{\prime}(x)=0$, we get a single root, $x=0$. Since the derivative function is continuous, it has a constant sign in the intervals $(-\infty ; 0)$ and $(0 ; \infty)$; based on $f^{\prime}(-1)=-1-\frac{3}{\sqrt{5}}<0$ and $f^{\prime}(1)=1+\frac{1}{\sqrt{5}}>0$, it is negative in $(-\infty ; 0)$ and positive in $(0 ; \infty)$. Therefore, $f(x)$ is decreasing in $(-\infty ; 0)$ and increasing in $(0 ; \infty)$. Thus, $f(x)$ has an absolute minimum at $x=0$.