: 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\frac{1}{2}-\frac{\varepsilon^{2}}{4 \cdot(1+\sqrt{2})} .
\end{aligned}
$$

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.