3. For positive real numbers $a, b, c$ it holds that

$$
a b+b c+c a=16, \quad a \geqq 3 .
$$

Find the smallest possible value of the expression $2 a+b+c$.
3. Let's modify the square of the expression $V=2a+b+c$, which is clearly positive. We will conveniently use the given relationship $ab+bc+ca=16$:

$$
\begin{aligned}
V^{2} & =(2a+b+c)^{2}=4a^{2}+b^{2}+c^{2}+4ab+4ac+2bc= \\
& =4a^{2}+b^{2}-2bc+c^{2}+4(ab+bc+ca)=4a^{2}+(b-c)^{2}+4 \cdot 16 .
\end{aligned}
$$

According to the problem, $a^{2} \geq 9$ and clearly $(b-c)^{2} \geq 0$, so $V^{2} \geq 4 \cdot 9 + 4 \cdot 16 = 100$. Therefore, $V \geq 10$.

Let's try to find numbers $a, b, c$ that satisfy the conditions and for which $V=10$. For the derived inequalities to become equalities, $b$ must equal $c$ and $a$ must equal 3, so then $ab+bc+ca=6b+b^{2}$. We need to find $b>0$ such that $6b+b^{2}=16$. This quadratic equation has roots $2$ and $-8$. The suitable triplet $(a, b, c)$ is thus $(3,2,2)$, and the minimum possible value of the given expression is indeed 10.

Another solution. From the given equation $ab+bc+ca=16$, we express $c=(16-ab)/(a+b)$ and substitute it into the expression $V$, whose minimum value we are looking for:

$$
\begin{aligned}
V=2a+b+c & =2a+b+\frac{16-ab}{a+b}= \\
& =a+b+\frac{16+a^{2}}{a+b} \geq 2 \sqrt{16+a^{2}} \geq 2 \sqrt{16+3^{2}}=10 .
\end{aligned}
$$

In the first inequality, we used the well-known inequality between the arithmetic and geometric means of two positive numbers $a+b$ and $(16+a^{2})/(a+b)$.

The triplet for which $V=10$ can be found as easily as in the first solution; the equality of the two averaged numbers occurs precisely when $a+b=\sqrt{16+a^{2}}$, which for $a=3$ immediately leads to the value $b=2$, so from the formula $c=(16-ab)/(a+b)$ we also get $c=2$.

Note. The key inequality $V \geq 10$ can also be obtained by the following manipulation:

$$
\begin{aligned}
& V=2a+b+c=(a+b)+(a+c) \geq \\
& \quad \geq 2 \sqrt{(a+b)(a+c)}=2 \sqrt{a^{2}+(ab+bc+ca)} \geq 2 \sqrt{3^{2}+16}=10
\end{aligned}
$$

in which we used the inequality between the arithmetic and geometric means of two positive numbers $a+b$ and $a+c$.

For a complete solution, award 6 points, of which 4 points for proving that $V \geq 10$, and 2 points for finding a corresponding triplet of numbers $a, b, c$ (even by guessing - it was not the task to find all such triplets), which satisfy the conditions and for which $V=10$.