Question 76
Find the smallest positive real number $\alpha$, such that
$$
\frac{x+y}{2} \geq \alpha\sqrt{xy} + (1-\alpha)\sqrt{\frac{x^2+y^2}{2}}
$$
for all positive real numbers $x$ and $y$.
$$
\frac{x+y}{2} \geq \alpha\sqrt{xy} + (1-\alpha)\sqrt{\frac{x^2+y^2}{2}}
$$
for all positive real numbers $x$ and $y$.