求 $x$ 的范围, 使得 $x^s+x^{\frac{1}{s}}+(1-x)^s+(1-x)^{\frac{1}{s}} \le 2$ 对所有的 $s\in(0,1)$成立.
aops(这个链接里没有正确的证明)
math.stackexchange.com/questions/3803042
用户Andreas的回答Note $\cos(50\pi (1-x)) = \cos(50\pi x)$ so indeed if we can prove that $f(x,k) = x^k+x^{\frac 1 k}+(1-x)^k+(1-x)^{\frac 1 k} \le 2$ for $x∈[0.25,0.75]$ and $k∈(0,1]$, we have proved a more general result than asked here.
If we fix $x$ and inspect $f(x,k)$ as a function of $k$ then it shows that for all $x$, $f(k)$ has only one minimum w.r.t. $k$, and the behavior is that $f(k=0) \to 2$, then $f(k)$ is falling monotonously with $k$ towards that minimum (interval 1), then $f(k)$ is rising monotonously (interval 2) until it reaches $f(k=1) = 2$.
To show this in the two intervals defined above, look at the derivatives. We have
$$
\partial f(x,k) / \partial k = \log(x) [x^k-\frac{1}{k^2}x^{\frac 1 k}] + \log(1-x) [ (1-x)^k-\frac{1}{k^2}(1-x)^{\frac 1 k} ]
$$
Consider interval 1. (The proof is yet given for this part.)
The two terms $x^k$ and $(1-x)^k$ are falling with $k$. So for establishing that no further solution $\partial f(x,k) / \partial k = 0$ exists, it is enough if we can show that also the terms ${k^2}x^{- \frac 1 k}$ and ${k^2}(1-x)^{-\frac 1 k}$ are falling with $k$. Let's again show this with calculus. Setting $g(k) = {k^2}x^{- \frac 1 k}$ gives $g'(k) = (2{k} + \log(x)) x^{- \frac 1 k}$ which is negative as long as $ k< - \frac12 \log(x)$. Likewise for the other term we require $ k< - \frac12 \log(1-x)$. Since we are in interval 1, we have (by inspection of the minimum which is at $x <0.5$) that the relevant (harder) condition is $ k< - \frac12 \log(1-x)$. However, this regime is actually larger than the required regime for interval 1, which can be seen by evaluating $\partial f(x,k) / \partial k $ at the limit $ k= - \frac12 \log(1-x)$ which shows that $\partial f(x,k) / \partial k > 0 $ for all $x$. This means that the condition $ k< - \frac12 \log(1-x)$ actually reaches into interval 2 where $f(k)$ is rising again, and we are safe. This proves interval 1.
Interval 2 should be proven similarly, I just didn't find time to do that yet. | 
