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The problem is trivial if at least one of $x$ or $y$ is greater than $1$. So all we need is to proof that $x^y+y^x > 1 \ \forall x,y \in (0,1)$.

Michael Rozenberg
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motoras
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    It's interesting to note that the limit doesn't exist as $(x,y) \to (0^+,0^+)$ as the path $x=0$ gives you $1$ and the path $y=x$ gives you $2$. – Kenny Lau Sep 14 '17 at 06:00
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    It may be of use to look at $1/x$ and $1/y$ with $x,y>1$ to Rephrase the condition as $$ \frac1{\sqrt[x]{y}}+\frac1{\sqrt[y]{x}}>1$$ – b00n heT Sep 14 '17 at 06:14
  • Well, there is the brute force attack using the second derivative test for a function of two variables, but it would be a bit messy. One would find that the minimum occurs when either $x=1$ or $y=1$. Hopefully, there is a slicker way to do it. – John Wayland Bales Sep 14 '17 at 06:20
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    Also: https://math.stackexchange.com/q/482549/42969, https://math.stackexchange.com/q/381090/42969 – all found instantly with Approach0 – Martin R Sep 14 '17 at 06:47
  • Thank you, I did search for the question on the site but I could not find it. I did not know about approach0. – motoras Sep 14 '17 at 13:00

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By Bernoulli $$\frac{1}{x^y}=\left(1+\frac{1}{x}-1\right)^y\leq1+y\left(\frac{1}{x}-1\right)=\frac{x+y-xy}{x}.$$ Thus, $$x^y\geq\frac{x}{x+y-xy}>\frac{x}{x+y}$$ and we are done.

Michael Rozenberg
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  • Isn't that essentially https://math.stackexchange.com/a/319620/42969 from the possible duplicate? – Martin R Sep 14 '17 at 07:17
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    I took the time to find 3 identical questions with answers. Why don't you (at least) use your golden badge to close it as a duplicate? As you know, you can always add better answers to the other questions. – Martin R Sep 14 '17 at 07:22
  • @Martin R Because I hope that my solution will give something new. I just like to prove inequalities more and more. Try to prove inequalities Martin R and I am sure that you'll understand me. – Michael Rozenberg Sep 14 '17 at 07:28
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    Apparently you still have not understood what "closing as a duplicate" is for and why it is a good thing to keep the site clean ... – Martin R Sep 14 '17 at 07:35
  • @Martin R I think that you still don't understand that besides site cleanliness there is math. If you wish to close it like duplicate then I will not protest. – Michael Rozenberg Sep 14 '17 at 07:56