1. Assume that each of the equations has at least one real solution.

For a quadratic equation $ax^2 + bx + c = 0$ to have real solutions, its discriminant must be nonnegative. The discriminant $\Delta$ of the quadratic equation $x^2 - 2bx + 2ac = 0$ is given by:
$$\Delta = (-2b)^2 - 4 \cdot 1 \cdot 2ac = 4b^2 - 8ac$$
For this equation to have real solutions, we must have:
$$4b^2 - 8ac \geq 0 \implies b^2 \geq 2ac$$

2. Apply the same reasoning to the other two equations:

For the equation $x^2 - 2cx + 2ab = 0$, the discriminant is:
$$\Delta = (-2c)^2 - 4 \cdot 1 \cdot 2ab = 4c^2 - 8ab$$
For this equation to have real solutions, we must have:
$$4c^2 - 8ab \geq 0 \implies c^2 \geq 2ab$$

For the equation $x^2 - 2ax + 2bc = 0$, the discriminant is:
$$\Delta = (-2a)^2 - 4 \cdot 1 \cdot 2bc = 4a^2 - 8bc$$
For this equation to have real solutions, we must have:
$$4a^2 - 8bc \geq 0 \implies a^2 \geq 2bc$$

3. Add the inequalities obtained:

$$b^2 \geq 2ac$$
$$c^2 \geq 2ab$$
$$a^2 \geq 2bc$$

Adding these three inequalities, we get:
$$a^2 + b^2 + c^2 \geq 2ac + 2ab + 2bc$$

4. Compare with the triangle inequality:

The triangle inequality states that the sum of the lengths of any two sides of a triangle must be greater than the length of the remaining side. This implies:
$$a + b > c, \quad b + c > a, \quad c + a > b$$
Squaring these inequalities and adding them, we get:
$$(a + b)^2 + (b + c)^2 + (c + a)^2 > c^2 + a^2 + b^2$$
Simplifying, we get:
$$2(a^2 + b^2 + c^2 + ab + bc + ca) > a^2 + b^2 + c^2$$
$$a^2 + b^2 + c^2 < 2(ab + bc + ca)$$

5. Derive the contradiction:

From the inequalities derived from the discriminants, we have:
$$a^2 + b^2 + c^2 \geq 2(ab + bc + ca)$$
But from the triangle inequality, we have:
$$a^2 + b^2 + c^2 < 2(ab + bc + ca)$$
This is a contradiction. Therefore, our initial assumption that each of the equations has at least one real solution must be false.

$\blacksquare$