Here is my question:

Prove that if $x,y,z>0$ then $$\frac{5x+3y+z}{5z+3y+x}+\frac{5y+3z+x}{5x+3z+y}+\frac{5z+3x+y}{5y+3x+z}\ge 3$$

Here is my attempt:

Let us assume $x>y>z$: \begin{align*} 4x+x+z &> 4z+x+z\\ 5x+z &> x+5z\\ 5x+3y+z &> x+3y+5z\\ \frac{5x+3y+z}{x+3y+5z} &> 1 \end{align*}

The question is given by my high school teacher as a challenge questions in my class.

I want to understand what is the difference between a cyclic inequality and symmetric inequality. Why $\frac{5x+3y+z}{5z+3y+x}+\frac{5y+3z+x}{5x+3z+y}+\frac{5z+3x+y}{5y+3x+z}\ge 3$ is not a symmetric inequality?

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    $\begingroup$ you can assume x>y>z, and try to attack it $\endgroup$ – Vivek Jan 23 '18 at 13:32
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    $\begingroup$ @Vivek how? can you please show me? $\endgroup$ – user521346 Jan 23 '18 at 13:32
  • $\begingroup$ How did you arrive at this equation ? Did you try to multiply with all the denominators ? $\endgroup$ – Peter Jan 23 '18 at 13:33
  • $\begingroup$ @Peter I tried, but it is not helping. $\endgroup$ – user521346 Jan 23 '18 at 13:34
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    $\begingroup$ Why is this inequality interesting? $\endgroup$ – Xander Henderson Mar 7 '18 at 1:04

It's $$\sum_{cyc}\left(\frac{5x+3y+z}{x+3y+5z}-1\right)\geq0$$ or $$\sum_{cyc}\frac{x-z}{x+3y+5z}\geq0$$ or $$\sum_{cyc}(x-z)(y+3z+5x)(z+3x+5y)\geq0$$ or $$\sum_{cyc}(x-z)(15x^2+5y^2+3z^2+28xy+14xz+16yz)\geq0$$ or $$\sum_{cyc}(15x^3+5x^2z+3x^2y+28x^2y+14x^2z+16xyz-$$ $$-15x^2z-5x^2y-3x^3-28xyz-14x^2y-16x^2z)\geq0$$ or $$12\sum_{cyc}(x^3+x^2y-x^2z-xyz)\geq0,$$ which is true by Rearrangement and AM-GM.

Also, we can use C-S.

Indeed, $$\sum_{cyc}\frac{5x+3y+z}{x+3y+5z}=\sum_{cyc}\frac{(5x+3y+z)^2}{(x+3y+5z)(5x+3y+z)}\geq$$ $$\geq\frac{\left(\sum\limits_{cyc}(5x+3y+z)\right)^2}{\sum\limits_{cyc}(x+3y+5z)(5x+3y+z)}\geq3,$$ where the last inequality it's just $$\sum_{cyc}(x-y)^2\geq0.$$ The inequality is named symmetric if it does not depend on all permutations of variables.

For example, the inequality $$a+b+c\geq3$$ is symmetric because for all permutations of $a$, $b$ and $c$ we obtain the same inequality: $$a+b+c\geq3;$$ $$a+c+b\geq3;$$ $$b+a+c\geq3;$$ $$b+c+a\geq3;$$ $$c+a+b\geq3$$ and $$c+b+a\geq3.$$ Easy to see that they are the same.

The inequality is named cyclic if it does not depend on all cyclic permutations of variables.

For example, the following inequality is cyclic. $$a^2b+b^2c+c^2a\geq3.$$ Cyclic permutations ($a\rightarrow b\rightarrow c\rightarrow a$) give: $$a^2b+b^2c+c^2a\geq3;$$ $$b^2c+c^2a+a^2b\geq3$$ and $$c^2a+a^2b+b^2c\geq3.$$ Easy to see that they are the same.

By the way, the last inequality is not symmetric because we can not say that it does not depend on all permutations of our variables.

For example, take the permutation $(a,b,c)\rightarrow(a,c,b)$.

By this permutation we get from $$a^2b+b^2c+c^2a\geq3$$ the following: $$a^2c+c^2b+b^2a\geq3,$$ which is something another.

  • $\begingroup$ @MichaelRozenberg what should be $\sum_{sym}\left(\frac{5x+3y+z}{x+3y+5z}\right)\geq3$ should be written as? $\endgroup$ – user521346 Jan 23 '18 at 19:33
  • $\begingroup$ It's exactly your inequality. I added something in my post. See now. $\endgroup$ – Michael Rozenberg Jan 23 '18 at 19:53
  • $\begingroup$ @MichaelRozenberg I got it... Thanks for your answer!!! There is a typing mistake in your answer, I don't have edit privilege so I can't correct it... Instead of $a+b+a\geq3$ it should be $c+b+a\geq3$ and instead of "inequality named cyclic/symmetric" it should be "inequality is named cyclic/symmetric" $\endgroup$ – user521346 Jan 23 '18 at 20:12
  • $\begingroup$ My English is very bad. I fixed. :) $\endgroup$ – Michael Rozenberg Jan 23 '18 at 20:16
  • $\begingroup$ @MichaelRozenberg as you said this inequality is symmetric, so can I use the technique "If both sides of an inequality are symmetric functions then there is no loss of generality in assuming $x>y>z$" ? Then I can add the three different inequalities $\frac{5x+3y+z}{x+3y+5z}> 1$ to get my resulting inequality. $\endgroup$ – user521346 Jan 23 '18 at 20:18

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