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.

I need to prove several inequalities trivially. (i.e. without using AM-GM, re-arrangement etc). I just keep hitting a blank. Could anyone help?

$$x^{4}+y^{4}+z^{4}\geq x^{2}yz+xy^{2}z+xyz^{2}$$

share|improve this question
    
$(x,y,z)=(1,1,1)$ –  pedja Apr 8 '12 at 10:36
    
@pedja: There are other solutions, e.g., $\langle x,y,z\rangle=\langle -1,1,1\rangle$. –  Brian M. Scott Apr 8 '12 at 10:36
    
Sorry I forgot to add that I have to prove this inequality holds true for all positive real numbers x,y,z. –  Roy Apr 8 '12 at 11:05
    
Using Muirhead's inequality these can be easily proven. –  karakfa Dec 21 '12 at 19:44
add comment

3 Answers

The righthand side is clearly $xyz(x+y+z)$. Assume that $x,y,z>0$. For fixed $x+y+z$, $xyz$ is maximized when $x=y=z$, and $x^4+y^4+z^4$ is minimized when $x=y=z$. This is easy to see if you can visualize the surfaces $x+y+z=k$, $xyz=k$, and $x^4+y^4+z^4=k$ for a positive constant $k$. Thus, for a fixed value of $x+y+z$, the worst case for the inequality is when $x=y=z$: that’s when the righthand side is biggest and the lefthand side is smallest. But when $x=y=z$, the two sides are clearly equal, so the inequality holds for all $x,y,z>0$. (From there, by the way, it’s easy to show that it holds for all real $x,y,z$.)

share|improve this answer
add comment

I'm guessing Sum-of-Squares representation would amount to a 'trivial' proof, right? How about expanding the following (this further proves that your inequality holds for all reals, not just positive ones..): $$\sum_{cyclic}(x^2 - y^2)^2 + \sum_{cyclic}x^2(y-z)^2 \ge 0$$

share|improve this answer
add comment

we know that, $$ \frac{x^{4}+y^{4}+z^{4}}{3}\geq (\frac{x+y+z}{3})^{4} $$

$$ x^{4}+y^{4}+z^{4}\geq \frac{1}{27}(x+y+z)^{4} $$

$\frac{x+y+z}{3} \geq (xyz)^{\frac{1}{3}}$

$(x+y+z)^3 \geq 27xyz$

hence $$ x^{4}+y^{4}+z^{4}\geq \frac{1}{27}(x+y+z)(27xyz) $$

$$x^{4}+y^{4}+z^{4}\geq x^{2}yz+xy^{2}z+xyz^{2}$$

share|improve this answer
add comment

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.