# Proving $x^x+y^y\ge\sqrt2$ when $x,y\in \mathbb R^+$ and $x+y=1$

Assuming $x,y\in \mathbb R^+$, $x+y=1$ how to prove $$x^x+y^y\ge\sqrt2$$thanks in advance

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Have you tried using $x\mapsto x^x+(1-x)^{1-x}$? –  xavierm02 Feb 3 '13 at 17:57
+1 nice question. What do you think about the proof? –  Babak S. Feb 3 '13 at 17:59
i use AM_GM inequality but it doesn't help until now –  Maisam Hedyelloo Feb 3 '13 at 18:04
However, it would be interesting to get an elementary solution based on AM-GM inequality... –  Chris's sis Feb 3 '13 at 18:09
@Chriss-sister:$x^x+y^y\ge2\sqrt{x^xy^y}$ then how complete it –  Maisam Hedyelloo Feb 3 '13 at 18:19

Let's consider the function $$f(x)=x \ln x +(1-x)\ln(1-x)$$ where $0<x<1$, and then $$f'(x)=\ln\frac{x}{1-x}$$ $$f'\left(\frac{1}{2}\right)=0$$ where $x_0=\frac{1}{2}$ is the point where the function reaches its minimum. Then $$f(x)\ge f\left(\frac{1}{2}\right)$$ that finally yields $$x \ln x +y \ln y \ge \ln\left(\frac{1}{2}\right)\tag1$$ Now, let's rewrite the left side inequality, use AM-GM inequality and then use $(1)$ $$e^{x \ln x}+e^{y\ln y}\ge 2 {\displaystyle e^{\displaystyle\frac{x \ln x+y \ln y}{2}}}\ge2 {\displaystyle e^{\displaystyle\frac{\ln(1/2)}{2}}}=\sqrt{2}$$

Chris.

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The AM-GM can be proven with Jensen, but this definitely a different approach. (+1) –  robjohn Feb 3 '13 at 20:59
@robjohn: thanks! When I realized that I need to use the derivatives, I tried to make things as easy as possible. This is what I got. :-) –  Chris's sis Feb 3 '13 at 21:03

Without Jensen: Lets minimize the function on in the positive quadrant. \begin{align} f(x)&=x^x+(1-x)^{(1-x)}\\ f^\prime(x)&=(1-x)^{(-x)} (-1+x) (1+\log(1-x))+x^x (1+\log(x)) \end{align} Equating to zero and solving. (You can also verify that the second derivative is positive) \begin{align} x&=0.5\\ f(x)&=(0.5)^{0.5}+(0.5)^{0.5}\\ f(x)&=2\times (0.5)^{0.5}\\ f(x)&=(4\times 0.5)^{0.5}\\ f(x)&=\sqrt{2}\\ \therefore~~~~~~ f(x)&\geq \sqrt{2}\quad \forall x\in \mathbb{R}^+ \end{align}

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Brute force works, too (+1) –  robjohn Feb 3 '13 at 21:00

HINT Prove that $x^x$ is convex in the interval $[0,1]$. Then use Jensen's inequality.

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Jensen is quite useful (+1) –  robjohn Feb 3 '13 at 21:02

Hint: Apply Jensen's Inequality on $f(x)=x^x$.

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Jensen is quite useful (+1) (Marvis beat you by a couple of minutes) –  robjohn Feb 3 '13 at 21:11
I know. I was working it out while typing out the answer and didn't realize it. Chris' sisters' method is more elegant though(in my opinion). –  Ishan Banerjee Feb 4 '13 at 12:18
Chris's sister's answer is nice, but when I saw the question, Jensen was the first thing that came to my mind. –  robjohn Feb 4 '13 at 15:50