2. Determine the minimum value of the expression

$$
V=x^{2}+\frac{2}{1+2 x^{2}}
$$

where $x$ is any real number. For which $x$ does the expression $V$ attain this value?
2. We will use the well-known inequality

$$
a+\frac{1}{a} \geqq 2
$$

which holds for any positive real number $a$, as it arises from the algebraic manipulation of the obvious inequality $(\sqrt{a}-1 / \sqrt{a})^{2} \geqq 0$. From this, it also follows that equality in it occurs precisely when $a=1$.

By a simple manipulation of the expression $V$, we get according to (1) for positive $a=\frac{1}{2}\left(2 x^{2}+1\right)$

$$
V=\left(\frac{2 x^{2}+1}{2}+\frac{2}{2 x^{2}+1}\right)-\frac{1}{2} \geqq 2-\frac{1}{2}=\frac{3}{2} .
$$

Equality in this inequality occurs precisely when $\frac{1}{2}\left(2 x^{2}+1\right)=1$, i.e., precisely when $x= \pm \frac{1}{2} \sqrt{2}$.

The smallest value of the expression $V$ is $\frac{3}{2}$, and this value is attained for $x= \pm \frac{1}{2} \sqrt{2}$.

Another solution. The manipulation

$$
\begin{aligned}
V & =x^{2}+\frac{2}{2 x^{2}+1}=\frac{2 x^{4}+x^{2}+2}{2 x^{2}+1}= \\
& =\frac{\frac{1}{2}\left(2 x^{2}-1\right)^{2}+3 x^{2}+\frac{3}{2}}{2 x^{2}+1}=\frac{1}{2} \cdot \frac{\left(2 x^{2}-1\right)^{2}}{2 x^{2}+1}+\frac{3}{2}
\end{aligned}
$$

which is correct for any real number $x$, shows that $V \geqq \frac{3}{2}$, and equality occurs precisely when $2 x^{2}-1=0$ or $x= \pm \frac{1}{2} \sqrt{2}$.

For a complete solution, award 6 points. Inequalities of type (1) (corresponding to the AM-GM inequality for two non-negative numbers) can be considered generally known and do not need to be proven. If both values of $x$ for which $V$ attains its minimum are not (correctly) determined, deduct 2 points.