How do we show that the $(1-\alpha)100\%$ confidence interval $(\bar{Y}-z\_{\alpha-\alpha\_1} \frac{\sigma}{\sqrt{n}}, \bar{Y}+z\_{\alpha\_1} \frac{\sigma}{\sqrt{n}})$ for the unknown normal mean $\mu$ has a minimal expected length when it is symmetric, that is, $\alpha\_1 =\frac{\alpha}{2}$

Intuitively the middle has the largest percentages. But how do we formally prove it?