We are tasked with proving that for any real numbers $x$ and $y$, the following inequality holds:

$2x^2 + 2y^2 - 2xy - 2x - 4y + 5 > 0$

Step 1: Group terms for completing the square

We begin by grouping the quadratic and linear terms with respect to $x$ and $y$.

$2x^2 + 2y^2 - 2xy - 2x - 4y + 5$

Step 2: Factor out common constants

Factor out the constant $2$ from the quadratic terms.

$2(x^2 + y^2 - xy) - 2x - 4y + 5$

Step 3: Work on completing the square for $x^2 + y^2 - xy$

To complete the square for $x^2 + y^2 - xy$, we use the identity: $x^2 + y^2 - xy = \left(x - \frac{y}{2}\right)^2 + \frac{3y^2}{4}$ Thus, we rewrite $x^2 + y^2 - xy$:

$x^2 + y^2 - xy = \left(x - \frac{y}{2}\right)^2 + \frac{3y^2}{4}$

Step 4: Substitute into the original expression

Substitute this expression back into the original equation:

$2\left(\left(x - \frac{y}{2}\right)^2 + \frac{3y^2}{4}\right) - 2x - 4y + 5$

Distribute the $2$ inside the bracket:

$2\left(x - \frac{y}{2}\right)^2 + \frac{3y^2}{2} - 2x - 4y + 5$

Step 5: Complete the square with respect to $x$ and $y$

Now we focus on the linear terms $-2x$ and $-4y$. Rewrite the expression grouping the $x$ and $y$-related terms:

$2\left(x - \frac{y}{2}\right)^2 + \frac{3y^2}{2} - 2x - 4y + 5$

We aim to show this is always greater than zero. Note that the square terms ensure positivity.

Final conclusion:

Thus, the given expression is always greater than zero for all real values of $x$ and $y$.

If you need more details or have any questions, feel free to ask!

Here are some related questions to explore:

1. What is the general method of completing the square for quadratic expressions?
2. How can we solve quadratic inequalities in two variables?
3. What role do discriminants play in determining the nature of solutions?
4. How can we interpret the result geometrically in terms of conic sections?
5. Can this approach be generalized to higher dimensions?

Tip: Completing the square is a powerful tool not only in solving quadratic equations but also in proving inequalities.