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p.s the original question is to prove that the graph of the function $f(x)=(1-x^p)^{\frac{1}{p}}$ concave down The inequality seems intuitive to me. because they subtract a small number from the left side. Right side is a number less than 1 and raised to positive powers then this number will become smaller and smaller. I have tried to use binomial theorem to expand the right side to see how things works. Then I get $-t^p\gt \sum_{k=1}^p (-t)^k$. Then I sum up the right side because it is a geometric series and I got $\frac{-t(1-(-t)^p)}{1+t}$, but here I kind of got confused and it seems what I did made the process messy. I thought there are better and smarter method.

Any hint or suggestion will be appreciated!

Userkkr
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2 Answers2

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By the strict convexity of the $p$-norm we have $$ (t^p+(1-t)^p)^{1/p}=\left\|\begin{matrix}t\\1-t\end{matrix}\right\|_p= \left\|t\begin{pmatrix}1\\0\end{pmatrix}+(1-t)\begin{pmatrix}0\\1\end{pmatrix}\right\|_p< t\left\|\begin{matrix}1\\0\end{matrix}\right\|_p+(1-t)\left\|\begin{matrix}0\\1\end{matrix}\right\|_p=t+1-t=1. $$

A.Γ.
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Fix any $t\in (0,1)$. If $p=1$, then we have equality. Take derivatives of both sides with respect to $p$; we have $-t^p\ln(t)$ on the left hand side and $(1-t)^p\ln(1-t)$ on the right hand side. Note that $-t^p\ln(t)$ is strictly positive for our range of $t$, whereas $(1-t)^p\ln(1-t)$ is strictly negative.

TomGrubb
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