Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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

4 Answers 4

up vote 5 down vote accepted

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.

share|improve this answer
    
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}

share|improve this answer
    
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.

share|improve this answer
    
Jensen is quite useful (+1) –  robjohn Feb 3 '13 at 21:02

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.