quick route to the inequality $(x+1)^p \le x^p + 1 + 2^p x (1+x^{p-2})$ for $x \ge 0$ and $p>2$

Question

Does anyone happen to know a quick way to show that $$ (x+1)^p \le x^p + 1 + 2^p x (1+x^{p-2}) \qquad (x \ge 0, \; p>2) \qquad ? $$ Differentiation is less than pleasant here, and the existence of standard inequalities involving $2^p$-type coefficients has me hoping this result can be obtained (possibly in a single line) from other known results.

Answer 1

The point is that either $x^{p-1} \ge x^{k}$ (for $x\ge 1$) or $x \ge x^{k}$ (for $x\le 1$) for all $1 \le k \le p-1,$ thus $x^k \le x^{p-1} + x.$ Actually, the $2^p$ can be improved to $2^{p}-2.$ Ignoring the $x^p + 1$ on both sides, you have in either case:

$$LHS = \sum_{k=1}^{p-1} x^k\binom{p}{k} \le (x^{p-1}+x)\cdot\sum_{k=1}^{p-1}\binom{p}{k} = (2^{p}-2)x(1+x^{p-2}).$$