Let $0<x\leq \frac{1}{2}$ define the function : $$f(x)=x^{\sqrt{2(1-x)}}+(1-x)^{\sqrt{2x}}$$

And let $f(x_0)$ be the minimum of the function on $(0,1/2)$

Then we have : $$f(x)+f(x_0)\leq 2$$

The maximum of the function is around $1.000150515\cdots$ and the minimum is around $0.9989495662\cdots$

I have tried derivative and Newton's method but it is not elegant so if you have a trick for this problem...

Thanks a lot .


1 Answer 1


COMMENT.-Since the derivative of $f(x)$ is complicated to solve to find $x_0$, we can treat the two functions in play separately. $$f_1(x)=x^{\sqrt{2(1-x)}}\\f_2(x)=(1-x)^{\sqrt{2x}}$$ Then with some values of $x\in[0,\frac12]$ we can verify that $f_1(x)$ is decreasing from $1$ to $\dfrac12$ and $f_2(x)$ is increasing from $0$ to $\dfrac12$.

This is enough to prove that $f(x)+f(x_0)\le2$


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