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Given $\forall x,y>0$
Prove that $x^y+y^x\ge1$

I have tried weighted inequalities and Jensen's but unfortunately ended up no where.
Please help me. (I know this is a basic inequality).

Michael Rozenberg
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Anish Ray
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2 Answers2

2

For $x\geq1$ or $y\geq1$ the inequality is obviously true.

But for $\{x,y\}\subset(0,1)$ easy to show that $x^y\geq\frac{x}{x+y}$ and we are done!

Michael Rozenberg
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  • Michael, it is easy to show that the inequality $x^y\ge \frac{x}{x+y}$ holds. And the conclusion is trivial thereafter. But, what inspired you to use this inequality? – Mark Viola Oct 20 '16 at 16:35
  • @Dr. MV 27 I think you solved one problem or maybe two in math. We do it by the following reasoning. What's enough to get for the proof? Since $\frac{x}{x+y}+\frac{y}{x+y}=1$, it's enough to prove that $x^y>\frac{y}{x+y}$, which is wrong. The rest it's for you. – Michael Rozenberg Oct 20 '16 at 17:10
  • First, I find your sarcastic tone highly offensive. You have not even come close to answering the question. What motivated you to pursue the solution using that specific inequality? There are a lot of ways to decompose $1$ into the sum of functions $f$ and $g$ that satisfy $f(x,y)=g(y,x)$. – Mark Viola Oct 20 '16 at 20:08
  • @Dr. MV It was a joke. Of course, there are very many ways. If to try, so we can find a right way. – Michael Rozenberg Oct 20 '16 at 20:45
  • That still didn't answer the question. I suspect that you'd seen this development or a similar one and simply applied it. And that's perfectly acceptable. All of us borrow what we've seen. ;-) -Mark – Mark Viola Oct 20 '16 at 21:09
  • @Dr. MV Now your tone is not good. I found this proof by myself, but indeed, around 30 years ago. By the way, in this forum there are very many my own inequalities, and copies of my solutions of another inequalities. I understand this situation. You are unfair to me. – Michael Rozenberg Oct 20 '16 at 21:34
  • Then just answer the simple question. What inspired you to appeal to this inequality? – Mark Viola Oct 20 '16 at 21:39
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Case I: $x$ and $y$ both greater than 1
$x^y, y^x > 1$
Hence, proved

Case II: x is 1 (without loss of generality)
$x^y$ = 1
$1 + y^x > 1$
Hence, proved

Case III: $x$ and $y$ both less than 1.
I can explain why this could be true informally. For both x and y less than 1, the power is actually a root, and a root of a number less than 1 will increase its value. Still, can't show that it will be at least one.

Starlight
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