A particle executes simple harmonic motion (amplitude = A) between $\mathrm{x}=-\mathrm{A}$  and $\mathrm{x}=+\mathrm{A}$. The time taken for it to go from 0 to A/2 is ${\mathrm{T}}_1$ and to go from A/2 to A is ${\mathrm{T}}_2$. Then 

Using  $\mathrm{x}=\mathrm{Asin\omega t}$
For  $\mathrm{x}=\mathrm{A}/2,{\mathrm{sin\omega T}}_1=1/2\Rightarrow {\mathrm{T}}_1=\frac{\mathrm{\pi }}{6\mathrm{\omega }}$
For  $\mathrm{x}=\mathrm{A},\text{\hspace{0.33em}}\mathrm{sin\omega }({\mathrm{T}}_1+{\mathrm{T}}_2)=1\Rightarrow {\mathrm{T}}_1+{\mathrm{T}}_2=\frac{\mathrm{\pi }}{2\mathrm{\omega }}$
$\Rightarrow {\mathrm{T}}_2=\frac{\mathrm{\pi }}{2\mathrm{\omega }}-{\mathrm{T}}_1=\frac{\mathrm{\pi }}{2\mathrm{\omega }}-\frac{\mathrm{\pi }}{6\mathrm{\omega }}=\frac{\mathrm{\pi }}{3\mathrm{\omega }}\mathrm{i}.\mathrm{e}.\text{\hspace{0.33em}}{\mathrm{T}}_1<{\mathrm{T}}_2$
Alternate method : In S.H.M., velocity of particle also oscillates simple harmonically. Speed is more near the mean position and less near the extreme position. Therefore the time taken for the particle to go from 0 to $\frac{\mathrm{A}}{2}$ will be less than the time taken to go from $\frac{\mathrm{A}}{2}$ to A. Hence  ${\mathrm{T}}_1<{\mathrm{T}}_2.$