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.
