Find the smallest positive real number $\alpha$, such that

$$
\frac{x+y}{2} \geq \alpha \sqrt{x y}+(1-\alpha) \sqrt{\frac{x^{2}+y^{2}}{2}}
$$

for all positive real numbers $x$ and $y$.
Let us prove that $\alpha=\frac{1}{2}$ works. Then the following inequality should hold for all positive real numbers $x$ and $y$:

$$
\begin{aligned}
& \frac{x+y}{2} \geq \frac{1}{2} \sqrt{x y}+\frac{1}{2} \sqrt{\frac{x^{2}+y^{2}}{2}} \\
\Longleftrightarrow & (x+y)^{2} \geq x y+\frac{x^{2}+y^{2}}{2}+2 \sqrt{x y \cdot \frac{x^{2}+y^{2}}{2}} \\
\Longleftrightarrow & (x+y)^{2} \geq 4 \sqrt{x y \cdot \frac{x^{2}+y^{2}}{2}} \\
\Longleftrightarrow & (x+y)^{4} \geq 8 x y\left(x^{2}+y^{2}\right) \\
\Longleftrightarrow & (x-y)^{4} \geq 0
\end{aligned}
$$

which is true, so we showed that $\alpha=\frac{1}{2}$ actually works.

Now it remains to show that $\alpha \geq \frac{1}{2}$. Let's consider $x=1+\varepsilon$ and $y=1-\varepsilon$ where $\varepsilon$ is a small positive number. Substituting these into the inequality, we get:

$$
\begin{aligned}
& \frac{(1+\varepsilon) + (1-\varepsilon)}{2} \geq \alpha \sqrt{(1+\varepsilon)(1-\varepsilon)} + (1-\alpha) \sqrt{\frac{(1+\varepsilon)^2 + (1-\varepsilon)^2}{2}} \\
\Longleftrightarrow & 1 \geq \alpha \sqrt{1 - \varepsilon^2} + (1-\alpha) \sqrt{1 + \varepsilon^2} \\
\Longleftrightarrow & 1 \geq \alpha \left(1 - \frac{\varepsilon^2}{2}\right) + (1-\alpha) \left(1 + \frac{\varepsilon^2}{2}\right) \\
\Longleftrightarrow & 1 \geq \alpha - \frac{\alpha \varepsilon^2}{2} + 1 - \alpha + \frac{(1-\alpha) \varepsilon^2}{2} \\
\Longleftrightarrow & 0 \geq -\frac{\alpha \varepsilon^2}{2} + \frac{(1-\alpha) \varepsilon^2}{2} \\
\Longleftrightarrow & 0 \geq \frac{\varepsilon^2}{2} (1 - 2\alpha) \\
\Longleftrightarrow & 0 \geq \varepsilon^2 (1 - 2\alpha)
\end{aligned}
$$

Since $\varepsilon^2 > 0$, we must have $1 - 2\alpha \leq 0$, which implies $\alpha \geq \frac{1}{2}$.

As $\varepsilon$ can be arbitrarily small, this expression can get arbitrarily close to $\frac{1}{2}$. This means that $\alpha < \frac{1}{2}$ cannot hold, as desired.