For positive real numbers $x,y,z$, Prove that,

$$\bigg(\frac{x^2+y^2+z^2}{x+y+z}\bigg)^{(x+y+z)}\geq x^xy^yz^z \geq \bigg(\frac{x+y+z}{3}\bigg)^{(x+y+z)}$$

I have no idea how to begin this? I am getting a hint for A.M. - G.M. equality but numbers don't seem to fit.

Thanks for you time.

  • $\begingroup$ Stacy: Where did you see this problem ? in a magazine ? a book or a homework. Which class you are taking that the Prof. assigned this problem? it's interesting. $\endgroup$ – DeepSea Aug 1 '18 at 17:11
  • $\begingroup$ @DeepSea This was problem in an assignment given by our professor. $\endgroup$ – Stacy Barrymore Aug 1 '18 at 17:15


Try using $x$ times $x$, $y$ times $y$ , $z$ times $z$.


Seemingly people downvote hints so updating the full solution.

Using A.M. - G.M. inequality,

$$\frac{x+x+..(x \;\Bbb{times})+y+y+..(y \;\Bbb{times})+z+z+..(z \;\Bbb{times})}{x+y+z}\geq (x^xy^yz^z)^{\frac{1}{x+y+Z}}$$

$$\bigg(\frac{x^2+y^2+z^2}{x+y+z}\bigg)^{(x+y+z)}\geq x^xy^yz^z $$

Using G.M. - H.M. inequality,

$$(x^xy^yz^z)^{\frac{1}{x+y+Z}} \geq \frac{x+y+z}{\frac{1}{x}+\frac{1}{x}+..(x \; \Bbb{times})+\frac{1}{y}+\frac{1}{y}+..(y \; \Bbb{times})+\frac{1}{z}+\frac{1}{z}+..(z \; \Bbb{times})}$$

$$ x^xy^yz^z \geq \bigg(\frac{x+y+z}{3}\bigg)^{(x+y+z)}$$

$$\bigg(\frac{x^2+y^2+z^2}{x+y+z}\bigg)^{(x+y+z)}\geq x^xy^yz^z \geq \bigg(\frac{x+y+z}{3}\bigg)^{(x+y+z)}$$

  • $\begingroup$ How did you type the answer that fast? You posted the answer within $2$ minutes the question was posted $\endgroup$ – tien lee Aug 1 '18 at 16:51
  • $\begingroup$ @tienlee copied authors notation and I work on a PC so. $\endgroup$ – prog_SAHIL Aug 1 '18 at 16:51
  • $\begingroup$ What does mean $\pi$ times? $\endgroup$ – Nosrati Aug 1 '18 at 16:52
  • $\begingroup$ Better to just refer to the Weighted AM-GM Inequality. As @user108128, "$\pi$ copies of $\pi$" does not really mean much. $\endgroup$ – Batominovski Aug 1 '18 at 16:53
  • $\begingroup$ @user108128 just to quantify the number of terms, nothing more. $\endgroup$ – prog_SAHIL Aug 1 '18 at 16:55

Since $\ln$ is a concave function we obtain: $$\sum_{cyc}\frac{x}{x+y+z}\ln{x}\leq\ln\left(\sum_{cyc}\frac{x}{x+y+z}\cdot x\right)=\ln\frac{x^2+y^2+z^2}{x+y+z},$$ which gives a left inequality.

The right inequality it's just Jensen for the convex function $f(x)=x\ln{x}:$ $$\frac{\sum\limits_{cyc}x\ln{x}}{3}\geq\frac{x+y+z}{3}\ln\frac{x+y+z}{3}.$$


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