Problem source

Show that $(a+b)^2\le 2a^2+2b^2\,$.
Find the stationary points on the curve  
$y=\big(a^2\cos^2\theta +b^2\sin^2\theta\big)^{\frac12}
+ \big(a^2\sin^2\theta +b^2\cos^2\theta\big)^{\frac12}\,$, where 
$a$ and $b$ are constants. State, with brief reasons, which points are maxima and which are minima. Hence prove that
\[
\vert a\vert +\vert b \vert 
\le \big(a^2\cos^2\theta +b^2\sin^2\theta\big)^{\frac12}
+ \big(a^2\sin^2\theta +b^2\cos^2\theta\big)^{\frac12}
\le \big(2a^2+2b^2\big)^{\frac12} \;.
\]

Solution source

\begin{align*}
&& 2a^2+2b^2 &= a^2 + b^2 + (a^2+b^2) \\
&&&\underbrace{\geq}_{AM-GM} a^2+b^2+2\sqrt{a^2b^2} \\
&&&= a^2+b^2 + 2|a||b| \\
&&&\geq a^2+b^2 + 2ab \\
&&&= (a+b)^2 
\end{align*}

Assume $a^2 \neq b^2$, otherwise the curve is a constant.

\begin{align*}
&& y & = \big(a^2\cos^2\theta +b^2\sin^2\theta\big)^{\frac12}
+ \big(a^2\sin^2\theta +b^2\cos^2\theta\big)^{\frac12}\\
&& \frac{\d y}{\d \theta} &= \tfrac12 \left (a^2\cos^2\theta +b^2\sin^2\theta \right)^{-\frac12} \cdot (2 \sin \theta \cos \theta (b^2 - a^2)) + \tfrac12 (a^2\sin^2\theta +b^2\cos^2\theta)^{-\frac12} \cdot (2 \sin \theta \cos \theta (a^2 - b^2) \\
&&&= \tfrac12\sin2 \theta (b^2 - a^2)  \left ( \left (a^2\cos^2\theta +b^2\sin^2\theta \right)^{-\frac12}  -  (a^2\sin^2\theta +b^2\cos^2\theta)^{-\frac12}\right) \\
\therefore  \frac{\d y}{\d x} = 0 \Rightarrow && \sin 2\theta = 0 & \text{ or } a^2\cos^2\theta +b^2\sin^2\theta = a^2\sin^2\theta +b^2\cos^2\theta \\
\Rightarrow && \theta &= 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \\
&& (a^2-b^2) \cos ^2\theta &= (a^2-b^2) \sin^2 \theta \\
\Rightarrow && \theta &= \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}
\end{align*}

WLOG $b^2 - a^2 > 0$, then the two parts of the derivative look like:

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){sin(2*(#1))};
    \def\functiong(#1){sqrt(1 + sin((#1))^2)-sqrt(1+cos(#1)^2)};
    \def\xl{-10};
    \def\xu{365};
    \def\yl{-1.1};
    \def\yu{1.1};
    
    % 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);
        
        \draw[thick, blue, smooth, domain=0:360, samples=100] 
            plot (\x, {\functionf(\x)});
        \draw[thick, red, smooth, domain=0:360, samples=100] 
            plot (\x, {\functiong(\x)});

        \draw[thick, green, smooth, domain=0:360, samples=1000] 
            plot (\x, {0.5*sign((\functiong(\x))*(\functionf(\x)))});
    \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}

And so $\frac{\pi}{4}, \frac{3\pi}{4}, \cdots$ are maxima, and the others minima. The maxima are where $\sin^2 \theta = \cos^2 \theta = \frac12$, so $y(\frac{\pi}{4}) = 2\left ( \frac{a^2+b^2}{2} \right)^{\frac12} = (2a^2+2b^2)^{\frac12}$ and the maxima are $\cos^2 \theta  = 1, \sin^2 \theta = 0$ and vice versa, ie $y = |a| + |b|$, therefore we obtain our desired result.