Is it possible to send a suitcase with an illegal measurements inside legal one The airport allows sending a suitcase if the sum of its length, width and height does not exceed a certain constant.
Question: Is it possible to send a suitcase with an illegal measurements by placing it inside one with legal ones?
 A: This is a classic tricky math puzzle. Here are two solutions, both of which can be found in Mathematical Mind-Benders, by Peter Winkler. The second is attributed to Andrei Storozhev.
Solution 1
Given any set $S$ of three dimensional space, let $S_\epsilon$ be the set of points which are within $\epsilon$ of some point of $S$. When $S$ is a box, $S_\epsilon$ is a larger box, but with rounded edges and corners. Specifically, you can show that when $S$ is a $l\times w\times h$ box, then
$$
\text{volume}(S_\epsilon) = \tfrac43\pi\epsilon^3 + (l+w+h)\pi\epsilon^2+2(lh+lw+hw)\epsilon+lhw 
$$
Furthermore, you can show that whenever $T\subseteq S$, then $T_\epsilon\subseteq S_\epsilon$. Therefore, if $S$ is the outer box, and $T$ is inner box, then $\text{volume}(T_\epsilon)\le \text{volume}(S_\epsilon)$.
Note that $\text{volume}(T_\epsilon)$  and $\text{volume}(S_\epsilon)$ is are cubic functions of $\epsilon$, with the same coefficient of $\epsilon^3$. Whichever has a larger coefficient of $\epsilon^2$ will be larger in the long run as $\epsilon\to +\infty$. Since $\text{volume}(S_\epsilon)$ is always greater, it must have the larger quadratic coefficient. But the quadratic coefficient is exactly the sum of the dimensions.
Solution 2
First, we prove the surface area of the inner box is smaller than that of the outer one. This can be seen by projecting the six faces of the inner box perpendicularly outwards onto the surface of the outer box. The six projections are distorted, with areas at least as large as their corresponding faces, and disjoint. Therefore, the outer box has more surface area than the inner box. This proves that $2(lh+lw+hw) > 2(l'h'+l'w' + h'w')$.
Furthermore, the diagonal of the larger box must longer than the diagonal of the shorter box. This can be seen by considering a sphere which circumscribes the outer box. The diameter of this sphere is the diagonal length of the outer box, and any two points in this sphere, including two opposite points of the inner box, have distance which is at most this diameter.  This proves $l^2+h^2+w^2\ge l'^2+w'^2+h'^2$.
Combining these two inequalities,
\begin{align}
(l+w+h)^2
&= l^2+w^2+h^2\;\;+2(lh+lw+hw)
\\&\ge l'^2+w'^2+h'^2+2(l'h'+l'w' + h'w')
\\&=(l'+w'+h')^2
\end{align}
A: No I don't think you could : 
Suppose that your suitcase is a rectangular parallelepiped, with dimension $l$, $w$, and $h$. Call $c$ the limit imposed by the Airline. You suitcase is illegal if
$$ l+w+h\geq c$$
If your suitcase can fit into another one of dimension $l'$, $w'$, $h'$, then necessarily
$$ \left\{ \begin{array}{l}
l\leq l'\\
w\leq w'\\
h\leq h'
\end{array}\right. \Rightarrow l'+w'+h'\geq l+w+h\geq c$$
Hence the new suitcase is also illegal. It does look like a trick question, in such case... wooosh
edit 1
If the problem is about the content of the suitcase, then you might as the cube maximize the volume for a given value of $l+w+h$.
