Verification to solution an inequality and proving another. I need solution verification an inequality, that I have solved because it seems too good to be true.
But first, I attempted this but couldn't complete:

Let $a$, $b$ and $c$ be the sides of a triangle with perimeter $3$. Prove that
$$
\sum_{cyc}{\frac{a^2}{a + 2\sqrt{b} - 1}} \geqslant 
\frac{ab^3 + bc^3 + ca^3 + 9abc}
{3(ab + bc + ca) - abc}
$$

Attempt:
By the constraint,
$$
\frac{ab^3 + bc^3 + ca^3 + 9abc}
{3(ab + bc + ca) - abc} =
\frac{
\sum_{cyc}{a^3b + 3a^2bc}
}{
\left(\sum_{cyc}{a^2(b + c)}\right) + 2abc
}
$$Which is by just writing $9$ as $3(a + b + c)$ and $3$ as $a + b + c$ in the $LHS$.
By $T_2$'s Lemma and then AM-GM Inequality,
$$
\sum_{cyc}{\frac{a^2}{a + 2\sqrt{b} - 1}}\geqslant \frac32
$$
Then what is left is
$$
\frac{a+b+c}2 \geqslant \frac{
\sum_{cyc}{a^3b + 3a^2bc}
}{
\left(\sum_{cyc}{a^2(b + c)}\right) + 2abc}
$$
$$
\Rightarrow\sum_{cyc}{a^3b + a^3c + a^2b^2 + a^2c^2 + 4a^2bc} \geqslant \sum_{cyc}{2ab^3 + 6a^2bc}
$$Then AM-GM again leaves us with
$$
\sum_{cyc}{a^3b}\geqslant \sum_{cyc}{ab^3}
$$Which means it's enough to prove
$$
\sum_{cyc}{a^2b - ab^2} \geqslant 0
$$But can't prove this. I have not used the fact that they are sides of triangle so maybe it is helpful somewhere. I wish alternative solutions to this inequality.
The second:

Let $x,y,z>0$ satisfy $xyz\geqslant1$. Prove that
$$
\frac {x^5 - x^2} {x^5 + y^2 + z^2} +
\frac {y^5 - y^2} {x^2 + y^5 + z^2} +
\frac {z^5 - z^2} {x^2 + y^2 + z^5} \geqslant 0
$$

I have proven the inequality but the solution seems too easy to me.
It is here:
$$
\sum_{cyc} {\frac{x^5 - x^2} {x^5 + y^2 + z^2}} \geqslant \sum_{cyc}{\frac{x^4 - x^2yz}{x^4 + y^3z +yz^3}} \geqslant \sum_{cyc}{\frac{x^4 - x^2yz}{x^4 + y^4 + z^4}} \geqslant 0
$$Which uses $$y^4 + z^4 \geqslant y^3z + yz^3 \Leftrightarrow (y - z)^2(y^2 + z^2 + yz)\geqslant 0\ \textrm{along with others}$$ and $$\sum_{cyc}{2x^4 + y^4 + z^4} \geqslant \sum_{cyc}{4x^2yz}$$Is this solution correct?
Thanks for comments and alternatives/extensions!
 A: The first inequality.
Since $$a+2\sqrt{b}-1=\frac{1}{3}(3a+6\sqrt{b}-a-b-c)=$$
$$=\frac{1}{3}\left(2a+2\sqrt{3b(a+b+c)}-b-c\right)>\frac{1}{3}(a+b-c)>0,$$ by AM-GM we obtain: $$\sum_{cyc}\frac{a^2}{a+2\sqrt{b}-1}\geq\sum_{cyc}\frac{a^2}{a+b+1-1}=\sum_{cyc}\frac{a^2}{a+b}$$ and it's enough to prove that:
$$\sum_{cyc}\frac{a^2}{a+b}\geq\frac{\sum\limits_{cyc}(a^3c+3abc)}{(a+b+c)(ab+ac+bc)-abc}$$ or
$$\sum_{cyc}\frac{a^2}{a+b}\geq\frac{\sum\limits_{cyc}(a^3c+3a^2bc)}{\prod\limits_{cyc}(a+b)}$$ or
$$\sum_{cyc}(a^3b+a^2b^2-2a^2bc)\geq0,$$ which is true by Rearrangement and SOS.
We'll prove the following general statement.

For positives $a$, $b$ and $c$ the triples $(a^2,b^2,c^2)$ and $(bc,ac,ab)$ have an opposite ordering.

Proof.
Since our claim is symmetric(is not changed after any permutations of $a$, $b$ and $c$),
we can assume that $a\geq b\geq c>0$.
Thus, $a^2\geq b^2\geq c^2$ and $bc\leq ac\leq ab$ and we are done.
By using of this statement and by Rearrangement we obtain: $$\sum_{cyc}a^3b=\sum_{cyc}(a^2\cdot ab)\geq \sum_{cyc}(a^2\cdot bc)=\sum_{cyc}a^2bc$$ and
$$\sum_{cyc}(a^2b^2-a^2bc)=\frac{1}{2}\sum_{cyc}c^2(a-b)^2\geq0.$$
Your solution of the second inequality is wrong because $x^4-x^2yz$ may be negative and you can not write $$\frac{x^4-x^2yz}{x^4+y^3z+yz^3}\geq\frac{x^4-x^2yz}{x^4+y^4+z^4}.$$
The second inequality we can prove by the following way.
Let $x=ka$, $y=kb$ and $z=kc$, where $k>0$ and $abc=1$.
Thus, $$k^3abc\geq1,$$ which gives $k\geq1.$
Now, by C-S we obtain: $$\sum_{cyc}\frac{x^5-x^2}{x^5+y^2+z^2}=\sum_{cyc}\frac{k^3a^5-a^2}{k^3a^5+b^2+c^2}=3+\sum_{cyc}\left(\frac{k^3a^5-a^2}{k^3a^5+b^2+c^2}-1\right)=$$
$$=3-\sum_{cyc}\frac{a^2+b^2+c^2}{k^3a^5+b^2+c^2}\geq3-\sum_{cyc}\frac{a^2+b^2+c^2}{a^5+b^2+c^2}=3-\sum_{cyc}\frac{bc(a^2+b^2+c^2)}{a^4+b^3c+bc^3}=$$
$$=3-\sum_{cyc}\frac{bc\left(1+\frac{b}{c}+\frac{c}{b}\right)(a^2+b^2+c^2)}{(a^4+b^3c+bc^3)\left(1+\frac{b}{c}+\frac{c}{b}\right)}\geq3-\sum_{cyc}\frac{bc\left(1+\frac{b}{c}+\frac{c}{b}\right)(a^2+b^2+c^2)}{(a^2+b^2+c^2)^2}=$$
$$=3-\sum_{cyc}\frac{2a^2+ab}{a^2+b^2+c^2}\geq3-\sum_{cyc}\frac{2a^2+a^2}{a^2+b^2+c^2}=0.$$
