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Suppose that $0 <a,b < 1$ and $a+b=1$

Today, I did some investigation of the expression $$a^{\frac{1}{b}}+b^{\frac{1}{a}}$$ and it seems that the maximum is at $a=b=\frac{1}{2}$ where the expression is equal to $\frac {1}{2}$.

I would like to conjecture that we have $$a^{\frac{1}{b}}+b^{\frac{1}{a}}\leq \frac {1}{2}$$

Is this true?

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  • $\begingroup$ @Macavity Where did you find greater value? $\endgroup$ – Grešnik Jun 24 at 16:15
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Consider the concave function $f(t) = \sqrt[t]{1-t}$ (shown below.) Using Jensen’s inequality,

$$f(a) + f(b) \leqslant 2f(\tfrac12) = \tfrac12$$

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To show $f$ is concave, it is enough to show $g=\log f$ is concave, as $t\mapsto e^t$ is convex and increasing. Perhaps the easiest way for this is to note the Taylor series for $x \in (0, 1)$ for $g= -\sum_{n \geqslant 0} \frac{x^n}{n+1} $ which implies all coefficients of $g’’$ are going to be negative as well $\implies g’’<0$.

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  • $\begingroup$ I didn´t know that concavity is saved with compositions. $\endgroup$ – Grešnik Jun 24 at 16:35
  • $\begingroup$ Not exactly, but there are some compositions with monotone functions which provide deterministic results. For e.g. if $h$ is convex, decreasing, and $g$ is concave, then $h(g)$ is convex. Helps identify convexity sometimes when derivatives are more difficult, as is the present case. $\endgroup$ – Macavity Jun 24 at 16:45
  • $\begingroup$ @Macavity I think, to prove by hand that $f$ is a concave function it's not so easy. $\endgroup$ – Michael Rozenberg Jun 24 at 17:29
  • $\begingroup$ @MichaelRozenberg Have added a note above which explicitly shows $\log f$ Is concave, which is enough to conclude $f$ is. $\endgroup$ – Macavity Jun 24 at 18:52

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