[quote=&quot;LucaMac&quot;:2kcrg639]I believe there's a mistake in the 5th exam paper's solution.[/quote:2kcrg639]

Absolutely right  <!-- s:oops: --><img src="{SMILIES_PATH}/icon_redface.gif" alt=":oops:" title="Embarassed" /><!-- s:oops: -->  <!-- s:oops: --><img src="{SMILIES_PATH}/icon_redface.gif" alt=":oops:" title="Embarassed" /><!-- s:oops: --> That inequality holds true only when [latex:2kcrg639]a^2\geq b^2[/latex:2kcrg639].

You also provided a nice proof that [latex:2kcrg639]u_0[/latex:2kcrg639] is not a WLM.  <!-- s:D --><img src="{SMILIES_PATH}/icon_biggrin.gif" alt=":D" title="Very Happy" /><!-- s:D -->  <!-- s:D --><img src="{SMILIES_PATH}/icon_biggrin.gif" alt=":D" title="Very Happy" /><!-- s:D --> 

A possible alternative approach is considering competitors of the form [latex:2kcrg639]u_\varepsilon&#40;x&#41;=\varepsilon\sin x\pm\varepsilon^2 v&#40;x&#41;[/latex:2kcrg639], where [latex:2kcrg639]v&#40;x&#41;[/latex:2kcrg639] is any function that is zero at the boundary and satisfies

[latex:2kcrg639]\displaystyle\int_0^\pi\left&#40;\cos x\cdot v&#39;&#40;x&#41;-\sin x\cdot v&#40;x&#41;\right&#41;\,dx\neq 0[/latex:2kcrg639].

It should not be difficult to show the existence of such a [latex:2kcrg639]v&#40;x&#41;[/latex:2kcrg639] (but one has to &quot;break the symmetry&quot;).