Find a point $X$, in the plane of regular pentagon $ABCDE$, that minimizes $\frac{XA+XB}{XC+XD+XE}$. Find such a point $X$, in the plane of the regular pentagon $ABCDE$, that the value of expression $$\frac{XA+XB}{XC+XD+XE}$$ is the lowest.


I tried using Ptolemy's theorem but don't know how to make use of inequalities it gives.

I'd be really grateful for any help :)
We use Ptolemy inequality : For an arbitrary quadrilateral $PRQS$ we have: $$PQ\cdot RS \leq PR\cdot QS+PS\cdot RQ$$


Let $AB = s$ and $AC=m$ and $XA =a$, $XB =b$... Now we use this inequality on $3$ different quadrilaterals:

$$AXBE:\;\;\;\;\; es\leq bs+am$$

$$ AXBC:\;\;\;\;\; cs\leq as+bm$$

$$ AXBD:\;\;\;\;\; ds\leq am+bm$$

Adding these three we get $$ s(c+d+e)\leq (a+b)(2m+s)$$so we have $$ {s\over 2m+s}\leq {a+b\over c+d+e}$$

So your expression is always $\geq  {s\over 2m+s}$ and this value is achieved if we put $X=A$ or $X=B$ or at any point on arc between $AB$ on circumcircle for $ABCDE$ 

(Remember that equality is true iff all inequalities become equalities and that is if $AXBC$, $AXBD$ and $AXBE$ are cyclic with that order of points on circle).
If $X$ is on $AB$, $AX+BX$ is minimum.  And if $X=A$ , $CX+DX+EX$ is maximum. So cesareo's answer follows.