Prove $x^2 + e^x \ge 1/e$
$x^{2}+e^{x} \geq 0+e^{x} >e^{-1}$ for $-1 < x<0$ since $e^{x}$ is an increasing function.
Prove $x^2 + e^x \ge 1/e$
$x^{2}+e^{x} \geq 0+e^{x} >e^{-1}$ for $-1 < x<0$ since $e^{x}$ is an increasing function.
$x^{2}+e^{x} \geq 0+e^{x} >e^{-1}$ for $-1 < x<0$ since $e^{x}$ is an increasing function.