Prove that 1/(2^(p-1)) ≤ x^p+ (1-x)^p ≤ 1 for x in [0,1 ]

Prove that $\dfrac{1}{2^{\,p-1}} \le x^{p}+\bigl(1-x\bigr)^{p} \le 1$ for $x\in\bigl[0,1\bigr]$ and $p\gt 1$.

Step 1: The function and its derivative

Fix $p\gt 1$ and set $f(x) = x^{p}+\bigl(1-x\bigr)^{p}$ on $\bigl[0,1\bigr]$. Then

\[ f'(x) = p\left[x^{\,p-1}-\bigl(1-x\bigr)^{p-1}\right]. \]

Step 2: The only critical point

Since $t\mapsto t^{\,p-1}$ is strictly increasing on $\bigl(0,\infty\bigr)$ for $p\gt 1$, the equation $x^{p-1} = \bigl(1-x\bigr)^{p-1}$ forces $x = 1-x$, i.e.

\[ x = \frac12. \]

Step 3: The function is convex

\[ f''(x) = p\bigl(p-1\bigr)\left[x^{\,p-2}+\bigl(1-x\bigr)^{p-2}\right] \gt 0 \qquad\text{on } \bigl(0,1\bigr), \] because $p\gt 1$ makes the factor $p(p-1)$ positive and both powers are positive (whatever the sign of the exponent $p-2$). So $f$ is strictly convex: the critical point is its minimum, and the maximum sits at an endpoint.

Step 4: The two bounds

\[ f\!\left(\frac12\right) = 2\cdot\frac{1}{2^{p}} = \frac{1}{2^{\,p-1}}, \qquad f(0) = f(1) = 1. \]

Hence $\dfrac{1}{2^{p-1}}\le f(x)\le1$ throughout $\bigl[0,1\bigr]$. Both bounds are attained: the lower one exactly at $x = \tfrac12$, the upper one exactly at the endpoints, so neither can be improved and neither inequality can be made strict.

Step 5: Verification

\[p = 3:\quad f(0) = 1,\ f(0.25) = 0.4375,\ f(0.5) = 0.25 = \tfrac{1}{2^{2}},\ f(0.75) = 0.4375,\ f(1) = 1 \;\;\checkmark\]

\[p = 1.5:\quad f(0.5) = 0.7071068 = 2^{-1/2}; \qquad f(0.25) = 0.7745190 \;\;\checkmark\]

\[\text{Symmetry } f(1-x) = f(x) \text{ alone forces the critical point to be at } x = \tfrac12 \;\;\checkmark\]