Bound on a normal quantile function $\phi(\Phi^{-1}(x)) \ge \sqrt{\frac{2}{\pi}} \min(x,1-x)$ In this post, the author has mentioned the following lower bound on the composition of a normal pdf and invers cdf
\begin{align}
\phi(\Phi^{-1}(x)) \ge \sqrt{\frac{2}{\pi}} \min(x,1-x), x\in (0,1).
\end{align}
Here $\phi$ is pdf of normal and $\Phi^{-1}$ is the invese CDF or quantile function.
I looked around and could not find such  bounds. How can something of this kind be shown?  
 A: By symmetry, it suffices to show that this holds for $x\in [0,1/2]$, i.e.
$$
\phi(\Phi^{-1}(x))\geq \sqrt{\frac{2}{\pi}}x.
$$
The strategy is to show that these functions agree on the endpoints, and then that the left-hand side is concave. This will imply the claim, as a concave function that agrees with a linear function at the endpoints must be at least as large between the endpoints.
For the first step, it is easy to see these are both $0$ at $x=0$ and $1/\sqrt{2\pi}$ at $x=1/2$, so it will suffice to show concavity. For $x\in (0,1)$, we may take the derivative twice of the left side, which by the Chain Rule, gives
\begin{equation}
\frac{d^2}{dx^2}[\phi(\Phi^{-1}(x))]=\frac{\phi''(\Phi^{-1}(x))\phi(\Phi^{-1}(x))-(\phi'(\Phi^{-1}(x)))^2}{(\phi(\Phi^{-1}(x)))^2},
\end{equation}
and it suffices to show that the numerator is nonpositive for all $x\in (0,1/2)$. But this is equivalent to the log-concavity of $\phi$. Recall that a function $f:\mathbb{R}\to \mathbb{R}_+$ is log-concave if $\log f$ is concave, or equivalently,
$$
f''f-(f')^2\leq 0.
$$
Here, $\log(\phi(x))=\log(\frac{1}{\sqrt{2\pi}})-x^2/2$, which is obviously concave (in fact, strictly concave). This implies the desired inequality.
