If $f(x)=(x-p)(x-q)(x-r)$, where p < q < r, are real numbers, then the application of Rolle's theorem on f leads to

$(p+q+r)^2>3(p q+q r+r p)$

We have,

$f(x) =(x-p)(x-q)(x-r)$

$\Rightarrow f(p) =0=f(q)=f(r)$

$\Rightarrow p, q$ and $r$ are three distinct real roots of $f(x)=0$

So, by Roll's theorem, $f'(x)$ has one real root in the interval $(p, q)$ and other in the interval $(q, r)$. Thus, $f'(x)$ has two distinct real roots.

Now,

$f(x)=(x-p)(x-q)(x-r)$

$\Rightarrow f(x)=x^3-x^2(p+q+r)+x(p q+q r+r p)-p q r$

$\Rightarrow f'(x)=3 x^2-2(p+q+r) x+(p q+q r+r p)$

As f'(x) has distinct real roots.

∴  $4(p+q+r)^2-12(p q+q r+r p)>0$

$\Rightarrow (p+q+r)^2>3(p q+q r+r p)$