# For $0 \le x \le 1 , p\gt 1$, prove $1 \over2^{p-1}$ $\le x^p +(1-x)^p \le 1$.

For $0 \le x \le 1 , p\gt 1$, prove $1 \over2^{p-1}$ $\le x^p +(1-x)^p \le 1$.

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One way to prove the first inequality is to use the fact that $x\mapsto x^p$ is a convex function to observe that $$\frac{1}{2^p}= \left(\frac{x+(1-x)}{2}\right)^p\leq \frac{x^p+(1-x)^p}{2}.$$ I prefer this method because it seems more insightful.
However, you can also do the straightforward thing and differentiate $x^p+(1-x)^p$, which for $p>1$ gives $px^{p-1}-p(1-x)^{p-1}$. This is zero iff $x=1-x$, i.e. $x=1/2$. Thus the extrema of $x^p+(1-x)^p$ occur at $0,1/2,1$ and you can observe that the value of $x^p+(1-x)^p$ at these points is $1,1/2^{p-1}$ and $1$ respectively.
From the answer here, we have for a real number $p \geq 1$ and complex numbers $\alpha, \beta$, $$|\alpha + \beta|^p \leq 2^{p-1}(|\alpha|^p + |\beta|^p)$$ In your case, take $\alpha = x \in [0,1]$ and $\beta = 1-x$ to obtain the lower bound on $x^p + (1-x)^p$. The upper bound is trivial since for $p>1$ and $x \in [0,1]$, we have $$x^p \leq x \,\, \text{ and } \,\, (1-x)^p \leq 1-x$$ Hence, we get $$x^p + (1-x)^p \leq x + 1-x = 1$$which gives us the upper bound.