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Let's prove that $\alpha =\frac{1}{2}$ works. Then the following inequality should hold for all positive real numbers $x$ and $y$: $$\begin{array}{rl}\frac{x+y}{2}& \ge \frac{1}{2}\sqrt{xy}+\frac{1}{2}\sqrt{\frac{x^2+y^2}{2}}\\ \iff (x+y{)}^2& \ge xy+\frac{x^2+y^2}{2}+2\sqrt{xy\cdot \frac{x^2+y^2}{2}}\\ \iff (x+y{)}^2& \ge 4\sqrt{xy\cdot \frac{x^2+y^2}{2}}\\ \iff (x+y{)}^4& \ge 8xy(x^2+y^2)\\ \iff (x-y{)}^4& \ge 0\end{array}$$ which is true, so we showed that $\alpha =\frac{1}{2}$ actually works.

Now it remains to show that $\alpha \ge \frac{1}{2}$. Let's consider $x=1+\epsilon$ and $y=1-\epsilon$ where $\epsilon <1$. Then the inequality becomes $$1\ge \alpha \sqrt{1-{\epsilon }^2}+(1-\alpha )\sqrt{1+{\epsilon }^2},\text{ i.e.}$$ $$\alpha \ge \frac{\sqrt{1+{\epsilon }^2}-1}{\sqrt{1+{\epsilon }^2}-\sqrt{1-{\epsilon }^2}}.$$ Notice that $$\begin{array}{rl}\frac{\sqrt{1+{\epsilon }^2}-1}{\sqrt{1+{\epsilon }^2}-\sqrt{1-{\epsilon }^2}}& =\frac{(\sqrt{1+{\epsilon }^2}-1)(\sqrt{1+{\epsilon }^2}+1)(\sqrt{1+{\epsilon }^2}+\sqrt{1-{\epsilon }^2})}{(\sqrt{1+{\epsilon }^2}-\sqrt{1-{\epsilon }^2})(\sqrt{1+{\epsilon }^2}+\sqrt{1-{\epsilon }^2})(\sqrt{1+{\epsilon }^2}+1)}\\ & =\frac{{\epsilon }^2(\sqrt{1+{\epsilon }^2}+\sqrt{1-{\epsilon }^2})}{2{\epsilon }^2(\sqrt{1+{\epsilon }^2}+1)}=\frac{\sqrt{1+{\epsilon }^2}+1-1+\sqrt{1-{\epsilon }^2}}{2(\sqrt{1+{\epsilon }^2}+1)}\\ & =\frac{1}{2}-\frac{1-\sqrt{1-{\epsilon }^2}}{2(\sqrt{1+{\epsilon }^2}+1)}=\frac{1}{2}-\frac{(1-\sqrt{1-{\epsilon }^2})(1+\sqrt{1-{\epsilon }^2})}{2(\sqrt{1+{\epsilon }^2}+1)(1+\sqrt{1-{\epsilon }^2})}\\ & =\frac{1}{2}-\frac{{\epsilon }^2}{2(\sqrt{1+{\epsilon }^2}+1)(1+\sqrt{1-{\epsilon }^2})}\\ & >\frac{1}{2}-\frac{{\epsilon }^2}{4(1+\sqrt{2})}.\end{array}$$ As $\epsilon$ 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.