An algebra problem by A Former Brilliant Member

The given expression can be written as :

$\left[\dfrac{2^{x^2}+1}{2}+\dfrac{2}{2^{x^2}+1}\right]-\dfrac{3}{2}+\dfrac{2^{x^2}}{2}$

Now, applying AM GM inequality inside the brackets we get :

$\dfrac{2^{x^2}+1}{2}+\dfrac{2}{2^{x^2}+1}-\dfrac{3}{2}+\dfrac{2^{x^2}}{2} \ge 2-\dfrac{3}{2}+\dfrac{2^{x^2}}{2}$

Minimum occurs at $x=0$

Note that $2^{x^2}$ also achieves its minimum at $x=0$

Hence, the minimum value of the function is :

$2-\dfrac{3}{2}+\dfrac{1}{2}=1$