P′² ≥ P·P″ for polynomials with real roots

Idea. The logarithmic derivative $\frac{P'}P=\sum\frac1{x-r_i}$ is a sum of terms whose derivatives are all negative.

Away from the roots. Let $P(x)=c\prod_{i=1}^n(x-r_i)$ with real $r_i$ and $c\ne0$. For $x\notin\{r_i\}$,
$$\frac{P'(x)}{P(x)}=\sum_{i=1}^n\frac1{x-r_i}.$$
Differentiating,
$$\frac{P''(x)P(x)-P'(x)^2}{P(x)^2}=-\sum_{i=1}^n\frac1{(x-r_i)^2}<0.$$
Multiplying by $P(x)^2>0$: $P'(x)^2-P(x)P''(x)=P(x)^2\sum_i\frac1{(x-r_i)^2}>0$.

At a root. If $P(x)=0$ then $P(x)P''(x)=0\le P'(x)^2$. $\blacksquare$

Example. For $P(x)=x^2-1$: $P'^2-PP''=2x^2+2>0$ ✓. For $P=x^2+1$ (no real roots): $4x^2-2(x^2+1)=2x^2-2<0$ at $x=0$, so the hypothesis cannot be dropped.

Lesson. The logarithmic derivative turns a product into a sum, and its derivative $-\sum\frac1{(x-r_i)^2}$ has a definite sign when the roots are real.