China Hong Kong Mathematical Olympiad

Firstly, when $\lambda = \frac{1}{2}$, we have
$$\begin{align*} \frac{a+b}{2} &\ge \frac{1}{2}\sqrt{ab} + \frac{1}{2}\sqrt{\frac{a^2+b^2}{2}} \\ \Leftrightarrow \quad (a+b)^2 &\ge ab + \frac{a^2+b^2}{2} + 2\sqrt{ab} \cdot \sqrt{\frac{a^2+b^2}{2}} \\ \Leftrightarrow \quad (a+b)^2 &\ge 2\sqrt{2ab(a^2+b^2)} \\ \Leftrightarrow \quad (a+b)^4 &\ge 8ab(a^2+b^2) \\ \Leftrightarrow \quad (a-b)^4 &\ge 0. \end{align*}$$

Secondly, suppose the inequality holds for all $a, b > 0$. Note that
$$\begin{align*} \frac{a+b}{2} &\ge \lambda\sqrt{ab} + (1-\lambda)\sqrt{\frac{a^2+b^2}{2}} \\ \Leftrightarrow \quad \sqrt{2}(\sqrt{a^2+b^2} - \sqrt{2ab})\lambda &\ge \sqrt{2(a^2+b^2)} - (a+b) \\ \Leftrightarrow \quad \frac{\sqrt{2}(a-b)^2}{\sqrt{a^2+b^2} + \sqrt{2ab}}\lambda &\ge \frac{(a-b)^2}{\sqrt{2(a^2+b^2)} + (a+b)}. \end{align*}$$

WLOG assume $a \neq b$. Then this holds if and only if
$$\lambda \geq \frac{\sqrt{a^2 + b^2} + \sqrt{2ab}}{\sqrt{2}(\sqrt{2(a^2 + b^2)} + (a+b))}.$$

We can rewrite this as
$$2\lambda \geq 1 - \frac{a+b-2\sqrt{ab}}{\sqrt{2(a^2+b^2)}+(a+b)} = 1 - \frac{(\sqrt{a}-\sqrt{b})^2}{\sqrt{2(a^2+b^2)}+(a+b)}.$$

Consider $a = 1$ and $b = 1 + \varepsilon$ where $\varepsilon \to 0$. It is clear that the right-hand side approaches 1. Therefore, in order that the inequality holds, we must have $2\lambda \geq 1$, which means $\lambda \geq \frac{1}{2}$.
Combining these, the smallest possible value of $\lambda$ is $\frac{1}{2}$.