The minimum value of the expression

Minimum value of 3x + 31 – x ; x ∈ R

Since arithmetic mean ≥ geometric mean of  3x and 31 – x

$\therefore \frac{3 x+3^{1-x}}{2} \geq \sqrt{3^{x} 3^{1-x}}$

i. e $\frac{3^{x}+3^{1-x}}{2} \geq \sqrt{3^{x} 33^{-x}}$

i. e $\frac{3^{x}+3^{1-x}}{2}=\sqrt{3}$

i. e $3^{x}+3^{1-x} \geq 2 \sqrt{3}$

i.e minimum possible value of $3^{x}+3^{1-x}$ is $2 \sqrt{3}$.