Prove inequality $2e^x>x^3+x^2$

For $x \leq 1$, notice that $$ x^3 + x^2 \leq x + 1 \leq e^x \leq 2e^x. $$

For $x \geq 1$, it suffices to prove that $f(x) := \log(2e^x) - \log(x^3 + x^2) > 0$. Differentiating twice, $$ f'(x) = 1 - \frac{2}{x} - \frac{1}{x+1}, \qquad f''(x) = \frac{2}{x^2} + \frac{1}{(x+1)^2} $$ This shows that $f$ is strictly convex and attains global minimum on $[1, \infty)$ at $x = 1+\sqrt{3}$. Now the conclusion follows from $$ f(1+\sqrt{3}) = 1+\sqrt{3} - 2\log(2+\sqrt{3}) \approx 0.098135 > 0. $$