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

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.

• 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. Commented Aug 1, 2018 at 17:11
• @DeepSea This was problem in an assignment given by our professor. Commented Aug 1, 2018 at 17:15

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 $f(x)=x\ln{x}:$ $$\frac{\sum\limits_{cyc}x\ln{x}}{3}\geq\frac{x+y+z}{3}\ln\frac{x+y+z}{3}.$$

HINT:

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

EDIT:

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,

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

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

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

• How did you type the answer that fast? You posted the answer within $2$ minutes the question was posted Commented Aug 1, 2018 at 16:51
• @tienlee copied authors notation and I work on a PC so. Commented Aug 1, 2018 at 16:51
• What does mean $\pi$ times? Commented Aug 1, 2018 at 16:52
• Better to just refer to the Weighted AM-GM Inequality. As @user108128, "$\pi$ copies of $\pi$" does not really mean much. Commented Aug 1, 2018 at 16:53
• @user108128 just to quantify the number of terms, nothing more. Commented Aug 1, 2018 at 16:55