Problem source

Sketch, without calculating the stationary points, the graph of
the function $\f(x)$ given by
\[
\f(x) = (x-p)(x-q)(x-r)\;,
\]
where $p < q < r$. By considering the quadratic  equation $\f'(x)=0$, or otherwise, show that
\[
(p+q+r)^2 > 3(qr+rp+pq)\;.
\]
By considering $(x^2+gx+h)(x-k)$, or otherwise, show that
$g^2>4h\,$ is a sufficient condition but not a necessary condition for the inequality
\[
(g-k)^2>3(h-gk)
\]
to hold.

Solution source

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){((#1)+.5)*((#1)-1)*((#1)-2.1)};
    \def\xl{-3};
    \def\xu{3};
    \def\yl{-10};
    \def\yu{10};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);

        \filldraw (-0.5, 0) circle (1.5pt) node[below]{$p$};
        \filldraw (1, 0) circle (1.5pt) node[below]{$q$};
        \filldraw (2.1, 0) circle (1.5pt) node[below]{$r$};
        
        \draw[thick, blue, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {\functionf(\x)});

        \node[blue, above, rotate=70] at (-1, {\functionf(-1)}) {\tiny $y = (x-p)(x-q)(x-r)$}; 
    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

Since there are two turning points the derivative (a quadratic) has two distinct real roots.
\begin{align*}
&& f'(x) &= 3x^2-2(p+q+r)x+(pq+qr+rp) \\
&& 0 &< \Delta = 4(p+q+r)^2 - 4\cdot 3(pq+qr+rp) \\
\Rightarrow && (p+q+r)^2 &>  3(pq+qr+rp)
\end{align*}

If $g^2 > 4h$ then $p(x) = (x^2+gx+h)(x-k)$ has at least 2 real roots (possibly one repeated, and in particular it has two turning point, ie

\begin{align*}
&& p'(x) &= (2x+g)(x-k)+(x^2+gx+h) \\
&&&= 3x^2+(2g-2k)x + (h-kg) \\
&& 0 &< \Delta = 4(g-k)^2 - 4\cdot 3 (h-gk) \\
\Rightarrow && (g-k)^2 &> 3(h-gk)
\end{align*}

Pick $g = h = 1$ and $k = 1000$ then $(-999)^2 > 0 > 3(1-1000)$ so it is sufficient but not necessary.