the inequality $2x^p - x^{p-1} - x + 2 \ge 0 \quad (x>0, \; p \ge 1)$ Calling the left member $f(x)$, I have $f'(x) = 2px^{p-1} - (p-1) x^{p-2} - 1 = 0$ as the equation for the critical point(s). There appears to be only one, in the interval $[0,1]$, but that's just an observation from plotting $f(x)$ for a few integer values of $p$. Looking for hints on how to get an actual proof of the inequality.
 A: We prove a stronger inequality (which implies obviously the given one).
$$2x^p - x^{p-1} - x + 2 \ge 1 \quad (x>0, \; p \ge 1)$$ In fact
$$2x^p - x^{p-1} - x + 1 \ge 0 \iff x^p+(x-1)(x^{p-1}-1)\ge0$$
The two functions $$f(x)=x^p\\g(x)=(x-1)(x^{p-1}-1)$$ are greater or equal than zero for $x\gt 0,\space p\gt1$ which prove the inequality. 
A: $$f(x)=x^{p-1}(x-2+x+1)-(x-2)=(x-2)(x^{p-1}-1)+(x+1)x^{p-1}$$
It's easy to note that $(x+1)x^{p-1}\gt 0$.
Can you show that $\forall x\gt 0$, either $(x-2)(x^{p-1}-1)\geq 0$ or $|(x+1)x^{p-1}|\geq |(x-2)(x^{p-1}-1)|$ is true?
This ensures that the sum $f(x)\geq 0$ is true $\forall~x\gt 0$
Hint: Consider what happens in the cases $x\geq 2$ and $x\lt 2$ for the above expression.
Hint #2: For $x\in (0,1]\cup [2,\infty)$, the first expression is $\geq 0$ (product terms have same sign) and for $(1,2)$, the only interval on which it fails, the second inequality holds (as shown in the comments).
A: Given that $p \geq 1$
For all $x \in (0,1)$ it easy to show that $2 -x^{p-1}-x \geq 0$.
For all $x \geq 1$ its easy to show that $2x^{p} -x^{p-1}-x \geq 0$ or equivalently $x^{p-1} (x-1) +x(x^{p-1}-1) \geq 0$.
Done.
