Verification to solution an inequality and proving another.

Question

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!

Answer 1

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: $$\sum_{cyc}\frac{x^5-x^2}{x^5+y^2+z^2}=\sum_{cyc}\frac{k^3a^5-a^2}{k^3a^5+b^2+c^2}=3+\sum_{cyc}\left(\frac{k^3a^5-a^2}{k^3a^5+b^2+c^2}-1\right)=$$ $$=3-\sum_{cyc}\frac{a^2+b^2+c^2}{k^3a^5+b^2+c^2}\geq3-\sum_{cyc}\frac{a^2+b^2+c^2}{a^5+b^2+c^2}=3-\sum_{cyc}\frac{bc(a^2+b^2+c^2)}{a^4+b^3c+bc^3}=$$ $$=3-\sum_{cyc}\frac{bc\left(1+\frac{b}{c}+\frac{c}{b}\right)(a^2+b^2+c^2)}{(a^4+b^3c+bc^3)\left(1+\frac{b}{c}+\frac{c}{b}\right)}\geq3-\sum_{cyc}\frac{bc\left(1+\frac{b}{c}+\frac{c}{b}\right)(a^2+b^2+c^2)}{(a^2+b^2+c^2)^2}=$$ $$=3-\sum_{cyc}\frac{2a^2+ab}{a^2+b^2+c^2}\geq3-\sum_{cyc}\frac{2a^2+a^2}{a^2+b^2+c^2}=0.$$