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Let $x,y$ be positive numbers. Prove that $x^x+y^y \ge x^y +y^x$.

This question appeared in Summer 1991 Russian Olympiad team test. Apparently, I tried to come up with different approach such as Jensen, Karamata's inequality and nothing works so far. I just need a discussion here. Hints are not necessary.

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  • $\begingroup$ :use $AM-GM$ inequality or define function $(f(x)=x^x)$ $\endgroup$ – M.H Jun 4 '13 at 11:03
  • $\begingroup$ Can you elaborate a bit more because I have used AM-GM but the difference in the power could be a major issue? $\endgroup$ – Viet Hoang Quoc Jun 4 '13 at 11:13
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    $\begingroup$ There is a long ML thread about this and related ones at artofproblemsolving.com/Forum/viewtopic.php?f=52&t=118722 $\endgroup$ – Wolfgang Oct 12 '13 at 16:41
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without loss of generality we only prove by $0\le y\le x\le 1$,let $$f(a)=a^{bx}-a^{by},x\ge a\ge y,1\ge bx-by\ge 0$$ then $$(a^{bx-by})'_{a}=\dfrac{bx-by}{a}\cdot a^{bx-by}>0,\Longrightarrow a^{bx-by}\ge y^{bx-by}\cdots (1)$$ since use $AM-GM$ inequality,we have $$y^{1+y-x}1^{x+xy-y^2-y}\le\left(\dfrac{x}{1+xy-y^2}\right)^{1+xy-y^2}\le x\cdots (2)$$ and $$f'(a)=\dfrac{bx}{a}\cdot a^{bx}-\dfrac{by}{a}\cdot a^{by}=\dfrac{ba^{by}}{a}(xa^{b(x-y)}-y)=$$ so use $(1),(2)$ we have $$f'(a)\ge ba^{by-1}(xy^{b(x-y)}-y)=ba^{by-1}y^{b(x-y)}(x-y^{1-b(x-y)})\ge 0 $$ so $$f(x)\ge f(y)\Longrightarrow x^{bx}+y^{by}\ge x^{by}+y^{bx}$$ let $b=1$

we have $$x^x+y^y\ge x^y+y^x$$

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  • $\begingroup$ The differentiation step does not look right to me, we have either $ (a^x)' = \ln (a) \times a^x $ or $(x^a)'=a \times x^{a-1}$. $\endgroup$ – Viet Hoang Quoc Jun 5 '13 at 8:15
  • $\begingroup$ I use $(x^a)'_{x}=a\times x^{a-1}$.. $\endgroup$ – math110 Sep 3 '13 at 0:43
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Let $x\geq y$.

We'll consider two cases.

  1. $x\geq1$.

Hence, $$x^x+y^y=x^y\left(x^{x-y}-1\right)+x^y+y^y\geq y^y\left(x^{x-y}-1\right)+x^y+y^y\geq$$ $$\geq y^y\left(y^{x-y}-1\right)+x^y+y^y\geq x^y+y^x.$$ 2. $1>x\geq y>0.$

Let $f(t)=t^x-t^y$, where $t\in[y,1).$

Thus, by Bernoulli we obtain: $$f'(t)=xt^{x-1}-yt^{y-1}=t^{y-1}\left(xt^{x-y}-y\right)\geq$$ $$\geq t^{y-1}\left(xy^{x-y}-y\right)=t^{y-1}y^{x-y}\left(x-y^{1-x+y}\right)=$$ $$=t^{y-1}y^{x-y}\left(x-(1+(y-1))^{1-x+y}\right)\geq$$ $$\geq t^{y-1}y^{x-y}\left(x-(1+(y-1)(1-x+y))\right)=t^{y-1}y^{x-y}y(x-y)\geq0,$$ which gives $$f(x)\geq f(y)$$ or $$x^x-x^y\geq y^x-y^y$$ or $$x^x+y^y\geq x^y+y^x$$ and we are done!

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Assume $x\ge y$ by symmetry. We want $$x^x-x^y\ge y^x-y^y.$$ i.e. $$x^y(x^{x-y}-1)\ge y^y(y^{x-y}-1),$$ which is then easy to show by listing all possible cases according to how $x,y$ compare to 1.

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Use induction, taking the base case x=0

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    $\begingroup$ This happens for all positive real number and induction clearly does not work nicely here. $\endgroup$ – Viet Hoang Quoc Jun 4 '13 at 11:29
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    $\begingroup$ This is wrong in the general case. @stancamp1, you may want to edit this answer or delete it in order to avoid downvotes. $\endgroup$ – DonAntonio Jun 4 '13 at 11:39

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