Prove double inequality $2^{1-p}\le x^p+(1-x)^p \le1$

Question

Prove that for any $p\ge1$ and $ x\in[0,1]$

The following inequalities are true: $$2^{1-p}\le x^p+(1-x)^p \le1$$ After I calculated the $\frac{d}{dp}$ of everything and got $-\frac{\ln{4}}{2^p}\le\ln{(x^{x^{p}}\cdot(1-x)^{(1-x)^{p}})}$ and I'm not sure what to do next, kinda seems that derivative may not be the case here but I don't see any place to use convexity or Taylor series expansion (which I currently have on my lectures).

Any help or hints will be appreciated.

Answer 1

Convexity of $(\cdot)^{p}$ gives $\left(\dfrac{1}{2}x+\dfrac{1}{2}(1-x)\right)^{p}\leq\dfrac{1}{2}x^{p}+\dfrac{1}{2}(1-x)^{p}$, so $2^{1-p}\leq x^{p}+(1-x)^{p}$.

On the other hand, we have $(a+b)^{1/p}\leq a^{1/p}+b^{1/p}$ for $p\geq 1$, so $(x^{p}+(1-x)^{p})^{1/p}\leq x+(1-x)=1$, so $x^{p}+(1-x)^{p}\leq 1$.

Answer 2

First, it should be $2^{1 - p}$, not $p - 1$. Second, the $p$-derivative is hard while $x$-derivative is not.

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Notice that if $f(x) = x^p + (1 - x)^p$, we have $f(0) = f(1) = 1$, while

$$\frac{df}{dx} = px^{p - 1} - p(1 - x)^{p - 1} = 0 \iff x = 1 - x \iff x = \frac 1 2$$

Now $f(1/2) = 2^{1 - p} \le 1$ and the bound follows.