I think so. The solution without typos : Note that $a^5 + b^5 \geq a^3b^2 + a^2b^3$ by Muirhead, rearrangement, AM - GM, ... so \begin{eqnarray*} \sum \frac{ab}{a^5 + b^5 + ab} &\leq& \sum \frac{ab}{a^3b^2 + a^2b^3 + ab} \\ \ &=& \sum \frac{1}{a^2b + ab^2 + 1} \\ \ &=& \sum \frac{abc}{a^2b + ab^2 + abc } \\ \ &=& \sum \frac{c}{a + b + c} \\ \ &=& 1. \end{eqnarray*} We're done !

One can easily show that $ x^5 + y^5 \geq x^2 y^3 + x^3 y^2$ for every x, y positive $ ab/(a^5+b^5+ab) \leq ab/(a^2 b^3 +a^3 b^2+ab) = 1/(a b^2+a^2 b+1) = 1/[ab(a+b+c)] =c/(a+b+c)$ since $ abc =1$ Similarly $ bc/(b^5+c^5+bc) \leq a/(a+b+c)$ and $ ca/(c^5+a^5+ca)\leq b/(a+b+c)$ By adding those three inequalities we get the result we are looking for. Cute no?

$ \frac {xy}{x^5 + y^5 + xy} = \frac {xyz}{x^5z + y^5z + xyz} = \frac {1}{x^5z + y^5z + 1}$ Now, $ \frac {1}{a+1} + \frac {1}{b+1} + \frac {1}{c+1} \implies a + b + c + 2 \le abc$ Thus, we have to show $ 2 + \sum {x^5y} = x^5y + x^5z + y^5x + y^5z + z^5x + z^5y + 2 \le (x^5 + y^5)(y^5 + z^5)(x^5 + z^5) = 2 + \sum {x^{10}y^5}$ But $ \sum{x^5y} = \sum{x^8y^4z^3}$, and we have Muirhead.

After homogenizing by turning the $ab$ terms into $a^2b^2c$ and etc and multiplying by 2, straight-up expansion gives the LHS to be $[11,2,2]+4[8,4,3]+[7,7,1]+2[7,6,2]+[5,5,5]$ and the RHS to be $[11,2,2]+2[10,5,0]+[7,7,1]+2[7,6,2]+2[8,4,3]+[5,5,5]$ Subtracting gives $RHS-LHS=2[10,5,0]-2[8,4,3]$ but $[10,5,0] \geq [8,4,3]$ by Muirhead so we're done.

1998,I proof: Suppose that $a, b, c > 0$ such that $abc = 1$ , $n$ be positive integer. Prove that \[ \frac{a^{n-1}b^{n-1}}{a^{n-1}b^{n-1} + a^{2n+1} + b^{2n+1}} + \frac{b^{n-1}c^{n-1}}{b^{n-1}c^{n-1} + b^{2n+1} + c^{2n+1}} \[\\ + \frac{c^{n-1}a^{n-1}}{c^{n-1}a^{n-1} + c^{2n+1} + a^{2n+1}} \leq 1. \] Another generalization ： Let $a,b,c $ be positive real numbers such that $abc=1 ,n\le-1 $ or $ n\ge2 ,$ Show that\[\frac{bc}{b^n+c^n+bc}+\frac{ca}{c^n+a^n+ca}+\frac{ab}{a^n+b^n+ab}\le1\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2835835 Similarly have Let $a,b,c>0 $ such that $abc=1 ,n\ge1 $ or $ n\le-2 ,$ Show that \[\frac{a^n}{a^n+b+c}+\frac{b^n}{b^n+c+a}+\frac{c^n}{c^n+a+b}\ge1\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2724850#p2724850

Arne wrote: Suppose that $a, b, c > 0$ such that $abc = 1$. Prove that \[ \frac{ab}{ab + a^5 + b^5} + \frac{bc}{bc + b^5 + c^5} + \frac{ca}{ca + c^5 + a^5} \leq 1. \] $a^5+b^5 = \frac{a^5}{5}+\frac{a^5}{5}+\frac{a^5}{5}+\frac{b^5}{5}+\frac{b^5}{5}+\frac{a^5}+{5}\frac{a^5}{5}+\frac{b^5}{5}+\frac{b^5}{5}\frac{b^5}{5} \ge a^3b^2+a^2b^3 $ Now , $\sum \frac{ab}{ab + a^5 + b^5} \le \sum \frac{1}{a^2b+ab^2+1} \le \sum \frac{abc}{ab(a+b+c} \le \sum \frac{c}{a+b+c)} = 1 \Box$

Notice that we can rewrite given $LHS$ as \[\sum_{cyc}{\frac{ab}{ab+a^5+b^5}}=\sum_{cyc}{\frac{1}{1+a^5c+b^5c}}.\] Then using supstiution $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ we get \[\sum_{cyc}{\frac{1}{1+a^5c+b^5c}}=\sum_{cyc}\frac{1}{1+\frac{x^4z}{y^5}+\frac{y^5}{z^4x}}\leq \sum_{cyc}\frac{1}{2\cdot \sqrt\frac{x^3}{z^3}+1}\] which we got by $AM-GM$. Then we have \[\sum_{cyc}\frac{1}{2\cdot \sqrt\frac{x^3}{z^3}+1}\leq\sum_{cyc}\frac{1}{2\cdot \sqrt\frac{x^3}{x^3}+1}=\sum_{cyc}\frac{1}{2+1}=\frac{1}{3}\cdot 3=1.\]Thats true because of rearrangement. $Q.E.D.$

@above: I don't see how you applied rearrangment in the final line. Also many solutions here use $a^5+b^5 \ge a^3b^2+b^2a^3$ (*). I have seen some times before the trick where you reduce an inequality to cylcic sum of $a/(a+b+c)$ equals 1, and if you have this trick in mind then it is quite easy to find that you need to prove (*), and then the problem is solved. But it seems to me a very unnatural thing to think of the trick first, and to me its not obvious at all that the trick is applicable here at first, so it does not come to mind. My question is then: how did you think to use this inequality (*)?

Given $x_1, x_2, x_3,....,x_{2015} >0$ and $x_1+x_2+x_3+....+x_{2015}=1$. Find minimum value of $P=(1+x_1^2)(1+x_2^2)(1+x_3^2)...(1+x_{2015}^2)$

@above: your post has nothing do with the problem in this thread. you should remove it i think.

Mewto55555 wrote: After homogenizing by turning the $ab$ terms into $a^2b^2c$ and etc and multiplying by 2, straight-up expansion gives the LHS to be $[11,2,2]+4[8,4,3]+[7,7,1]+2[7,6,2]+[5,5,5]$ and the RHS to be $[11,2,2]+2[10,5,0]+[7,7,1]+2[7,6,2]+2[8,4,3]+[5,5,5]$ Subtracting gives $RHS-LHS=2[10,5,0]-2[8,4,3]$ but $[10,5,0] \geq [8,4,3]$ by Muirhead so we're done. I don't know what he mean by $[11,2,2]+4[8,4,3]+[7,7,1]+2[7,6,2]+[5,5,5]$ and $[11,2,2]+2[10,5,0]+[7,7,1]+2[7,6,2]+2[8,4,3]+[5,5,5]$ Can someone explain ?

velraman wrote: Mewto55555 wrote: After homogenizing by turning the $ab$ terms into $a^2b^2c$ and etc and multiplying by 2, straight-up expansion gives the LHS to be $[11,2,2]+4[8,4,3]+[7,7,1]+2[7,6,2]+[5,5,5]$ and the RHS to be $[11,2,2]+2[10,5,0]+[7,7,1]+2[7,6,2]+2[8,4,3]+[5,5,5]$ Subtracting gives $RHS-LHS=2[10,5,0]-2[8,4,3]$ but $[10,5,0] \geq [8,4,3]$ by Muirhead so we're done. I don't know what he mean by $[11,2,2]+4[8,4,3]+[7,7,1]+2[7,6,2]+[5,5,5]$ and $[11,2,2]+2[10,5,0]+[7,7,1]+2[7,6,2]+2[8,4,3]+[5,5,5]$ Can someone explain ? He is probably referring to the coefficients, and the ordered triplet represents the powers of a,b, and c respectively.

My Solution : Observe that $(x^3-y^3)(x^2-y^2) \ge 0 \implies x^5 + y^5 \ge x^2y^2(x+y) \ \forall x,y \in \mathbb{R^+}$. Thus, $\sum_{cyc} \dfrac {ab}{a^5+b^5+ab} \le \sum_{cyc} \dfrac{ab}{a^2b^2(a+b)+ab} = \sum_{cyc} \dfrac{c}{a+b+c} = 1$.

\[(ab+a^5+b^5)\left(\frac{c^4}{ab} + \frac1a + \frac1b\right)\ge (a^2+b^2 + c^2)^2\]\[\iff \frac{ab}{ab+a^5+b^5} \le \frac{c^4 + a + b}{(a^2 + b^2 + c^2)^2}\]Summing up similar inequalities. \[\text{LHS} \le \frac{\sum a^4 +2\sum a }{(a^2 + b^2 + c^2)^2}\]So, it suffices to prove that \begin{align*} \frac{\sum a^4 +2\sum a }{(a^2 + b^2 + c^2)^2}&\le 1\\ \iff a^2b^2 + b^2c^2 + c^2a^2 &\ge a + b + c\\ &= abc(a+b+c)\\ \iff \sum a^2(b-c)^2 &\ge 0 \end{align*}

Nice one Vrangr

observe that $$x^{5}+y^{5} = (x+y)(x^{4}-x^{3}y+x^{2}y^{2}-xy^{3}+y^{4})$$$$=(x+y)(x^{3}(x-y) -y^{3}(x-y) +x^{2}y^{2}) = (x+y)((x-y)^{2}(x^{2}+y^{2}+xy)+x^{2}y^{2})\ge (x+y)(x^{2}y^{2})$$since $(x-y)^{2}\ge0$ $$\Rightarrow \sum_{cyc}\frac{xy}{x^{5}+y^{5}+xy}\le \sum\frac{xy}{(x+y)(x^{2}y^{2})+xy}= \sum\frac{1}{(x+y)xy+1}$$$$=\sum\frac{xyz}{(x+y)xy+xyz}= \sum\frac{z}{x+y+z}=1$$And we are done!

Arne wrote: Suppose that $a, b, c > 0$ such that $abc = 1$. Prove that \[ \frac{ab}{ab + a^5 + b^5} + \frac{bc}{bc + b^5 + c^5} + \frac{ca}{ca + c^5 + a^5} \leq 1. \] $$\sum\frac{ab}{ab+a^5+b^5}=\sum\frac{a^2b^2c}{a^2b^2c+a^5+b^5}\leq\sum\frac{a^2b^2c}{a^2b^2c+a^3b^2+a^2b^3}=\sum\frac{c}{a+b+c}=1$$

Observe that (By Muirhead) $a^5+b^5\ge a^2b^2(a+b)$. We have \begin{align*} \sum_{cyc}\frac{ab}{a^5+ab+b^5}&\le\sum_{cyc}\frac{ab}{a^2b^2(a+b)+ab} \\ &=\sum_{cyc}\frac1{ab(a+b)+1} \\ &=\sum_{cyc}\frac1{ab(a+b)+abc} \\ &=\sum_{cyc}\frac{1}{ab(a+b+c)} \\ &=\sum_{cyc}\frac c{a+b+c} \\ &=1 \end{align*} Very similar to a problem from USAMO 1998.

Arne wrote: Suppose that $a, b, c > 0$ such that $abc = 1$. Prove that \[ \frac{ab}{ab + a^5 + b^5} + \frac{bc}{bc + b^5 + c^5} + \frac{ca}{ca + c^5 + a^5} \leq 1. \] Let $a,b,c>0$ with $abc=1.$ Establish that $$\frac{1}{a^5+b^5+1}+\frac{1}{b^5+c^5+1}+\frac{1}{c^5+a^5+1}\leq 1.$$$$\iff$$$$\frac{1}{a^3+b^3+1} +\frac{1}{b^3+c^3+1} +\frac{1}{c^3+a^3+1} \leq 1$$h h h

By Muirhead’s inequality, $b^5 + c^5 \ge b^3c^2 + b^2c^3$. Hence, $$\sum_{cyc} \frac{ab}{ab+a^5+b^5} \le \sum_{cyc} \frac{ab}{ab+a^3b^2 + a^2b^3} = \sum_{cyc} \frac{1}{1+a^2b+b^2a} = \sum_{cyc} \frac{c}{c+a+b} = 1$$as desired.

levinhkiet wrote: Given $x_1, x_2, x_3,....,x_{2015} >0$ and $x_1+x_2+x_3+....+x_{2015}=1$. Find minimum value of $P=(1+x_1^2)(1+x_2^2)(1+x_3^2)...(1+x_{2015}^2)$ easy the answer is $\frac{2016}{2015}$

WLOG let $a$$\leq$$ b$$ \leq c$. So we have, from rearrangement inequality, $\begin{bmatrix} a^3 & b^3 \\ a^2 & b^2 \end{bmatrix} \geq \begin{bmatrix} a^3 & b^3 \\ b^2 & a^3 \end{bmatrix}$. Which in turn gives , $a^5+b^5 \geq a^{3}b^{2} + a^{2}b^{3} $ $\frac {ab}{a^5+b^5 +ab} $$\leq$$ \frac {ab}{ a^{3}b^{2} + a^{2}b^{3}+ab}$ Then it follows

Note that $\frac{xy}{x^5+ xy+y^5}=\frac{1}{z(x^5+y^5)+1} \leq \frac{1}{z(x^4y+xy^4)+1}=\frac{1}{(x^3+y^3+1)} \leq \frac{1}{xy(x+y)+1}=\frac{z}{x+y+z}$ and the result follows by summing this cyclically $\Box$

$a^5+b^5\ge a^4b+ab^4$ and $a^3+b^3\ge a^2b+ab^2$ are direct by Rearrangement, so we have: $\sum_{\text{cyc}}\frac{ab}{ab+a^5+b^5}\le\sum_{\text{cyc}}\frac1{1+a^3+b^3}\le\sum_{\text{cyc}}\frac c{c+a^2bc+ab^2c}=\sum_{\text{cyc}}\frac a{a+b+c}=1$

For positive reals $x,y,z$ with $xyz=1$, prove or disprove that: $$\frac1{x+y+1}+\frac1{y+z+1}+\frac1{z+x+1}\le1.$$

The classic Muirhead for denominator inequalities! We have that $\sum_{\text{cyc}} \frac{ab}{ab+a^5+b^5}\leq \sum_{\text{cyc}} \frac{ab}{ab+a^3b^2+a^2b^3} = \sum_{\text{cyc}} \frac{1}{abc+a^2b+ab^2}=\frac{a+b+c}{a+b+c}=1.$ Remarks: In the beginning, I tried to use AM-GM directly on the denominator to get that I wanted to show that $\sum_{cyc} \frac{1}{ab}\leq 3\implies a+b+c\leq 3,$ but obviously the inequality was too weak.

\begin{align*} \sum_{\text{cyc}} \dfrac{ab}{{\color{red}a^5+b^5}+ab} &\le \sum_{\text{cyc}} \dfrac{ab}{{\color{blue}a^3b^2 + a^2b^3} + ab} \\ &= \sum_{\text{cyc}} \frac{ab}{ab(a^2b + ab^2 + {\color{red}1})}\\ &= \sum_{\text{cyc}} \frac{1}{a^2b + ab^2 + {\color{blue}abc}} \\ &= \sum_{\text{cyc}} \frac{1}{{\color{red}ab}(a+b+c)} \\ &= \sum_{\text{cyc}} \frac{\color{blue}c}{a+b+c}\\ &= 1 \end{align*}

By Muirhead's Theorem: $a^5+b^5 \geq a^2b^3+a^3b^2$ $\sum_{\text{cyc}} \frac{ab}{a^5+b^5+ab}\leq \sum_{\text{cyc}} \frac{ab}{a^3b^2+a^2b^3+ab} = \sum_{\text{cyc}} \frac{1}{a^2b+ab^2+1}=\sum_{\text{cyc}}\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1$.$\blacksquare$

Suppose that $a, b, c > 0$ such that $abc = 1$. Prove that$$\frac{ab}{2ab + a^5 + b^5} + \frac{bc}{2bc + b^5 + c^5} + \frac{ca}{2ca + c^5 + a^5} \leq \frac{3}{4}$$$$\frac{ab}{ab + 2a^5 + b^5} + \frac{bc}{bc +2b^5 + c^5} + \frac{ca}{ca + 2c^5 + a^5} \leq \frac{3}{4}$$$$ \frac{ab}{kab + a^5 + b^5} + \frac{bc}{kbc + b^5 + c^5} + \frac{ca}{kca + c^5 + a^5}\leq \frac{3}{k+2}$$Where $k\geq 1.$

Arne wrote: Suppose that $a, b, c > 0$ such that $abc = 1$. Prove that \[ \frac{ab}{ab + a^5 + b^5} + \frac{bc}{bc + b^5 + c^5} + \frac{ca}{ca + c^5 + a^5} \leq 1. \] $(a^3-b^3)(a^2-b^2) \geq 0 \implies a^5+b^5 \geq a^2b^2(a+b),$ equality when $a=b$. So, $\frac{ab}{a^5+b^5+ab} \leq \frac{ab}{a^2b^2(a+b)+ab} = \frac{1}{ab(a+b)+1} =\frac{abc}{ab(a+b+c)} =\frac{c}{a+b+c}.$ Similarly, $\frac{bc}{b^5+c^5+bc} \leq \frac{a}{a+b+c}$ and $\frac{ca}{c^5+a^5+ca} \leq \frac{b}{a+b+c},$ equality when $a=b=c=1.$ The claim follows adding these three. $\square$

Same as above solutions; posting for storage. By Muirhead, $a^5+b^5\ge a^3b^2+b^3a^2.$ Hence, $$\sum_{\text{cyc}}\frac{ab}{ab+a^5+b^5}\le\sum_{\text{cyc}}\frac{ab}{ab+a^3b^2+b^3a^2}=\sum_{\text{cyc}}\frac{1}{\frac{a}{c}+\frac{b}{c}+\frac{c}{c}}=\sum_{\text{cyc}}\frac{c}{a+b+c}=1.$$$\square$

Every solution is going to look the same I guess. We first show that $a^5+b^5 \ge a^3b^2++b^3a^2$. This will be just expanding and applying AM-GM inequality. \begin{align*} a^5+b^5 = (a+b)(a^4-a^3b+a^2b^2-ab^3+b^4) \ge (a+b)(2a^2b^2+a^2b^2-ab(a^2+b^2)) \ge (a+b)(3a^2b^2-2a^2b^2) = a^3b^2+b^3a^2 \end{align*}Returning to the problem: \begin{align*} \sum_{\text{cyc}}\frac{ab}{ab+a^5+b^5} = \sum_{\text{cyc}}\frac{1}{1+c(a^5+b^5)} \le \sum_{\text{cyc}}\frac{1}{1+c \cdot a^2b^2(a+b)} = \sum_{\text{cyc}}\frac{1}{1+ab(a+b)} = \sum_{\text{cyc}}\frac{c}{a+b+c} =1 \end{align*}which means we are done. $\blacksquare$

jasperE3 wrote: For positive reals $x,y,z$ with $xyz=1$, prove or disprove that: $$\frac1{x+y+1}+\frac1{y+z+1}+\frac1{z+x+1}\le1.$$ Interesting. You can easily prove this result is true from straight-up expansion and then Muirhead. Note that this is actually equivalent to the original problem. From the substitution $x=a^5b$, $y=b^5c$, etc, the LHS turns into $\sum_{cyc}{\frac{1}{ab^5 + bc^5 + 1}}$, and dividing both top and bottom by $abc=1$ gives the original inequality. I would be curious to see if anyone has a solution from Holder's/Cauchy's/Titu's (from $\sum_{cyc}{\frac{a^5+b^5}{ab+a^5+b^5}} \ge 2$.) I tried but didn't get anywhere.

By Power-Mean Inequality, we obtain $$\sqrt[5]{\frac{a^5+b^5}{2}} \geq \frac{a+b}{2}$$\begin{align*} \implies \frac{a^5+b^5}{2} &\geq \left(\frac{a+b}{2}\right)^5 \\ &\geq \left(\frac{a+b}{2}\right)^4 \cdot \frac{a+b}{2} \\ & \geq a^2b^2\left(\frac{a+b}{2}\right) \text{[using AM-GM]}\\ \end{align*}$$\therefore \boxed{a^5+b^5\geq a^2b^2(a+b)}.$$ Hence, \begin{align*} \sum_{cyc} {\frac{ab}{a^5+b^5+ab}} &\leq \sum_{cyc} {\frac{ab}{a^2b^2(a+b)+ab}} \\ &= \sum_{cyc} {\frac{1}{ab(a+b)+1}} \\ &= \sum_{cyc} {\frac{c}{a+b+c}} \text{[using $abc=1$]} \\ &= 1. \end{align*}$QED.$

Solved with CT17. Notice that $(b^3-c^3)(b^2-c^2) \ge 0 \implies b^5+c^5 \ge b^3c^2+b^2c^3$, so we have \[\sum_\text{cyc} \frac{bc}{bc+b^5+c^5} \le \sum_\text{cyc} \frac{bc}{bc+b^3c^2+b^2c^3}=\sum_\text{cyc} \frac{\frac{1}{a}}{\frac{1}{a}+\frac{b}{a^2}+\frac{c}{a^2}}=\sum_\text{cyc} \frac{a}{a+b+c}=1,\]as desired. $\square$

Version 1 Let $a,b,c$ be positive reals such that $abc=2$. Then prove that $$\sum_{cyc}{\dfrac{a^2b^2}{2\left(a^2b^2\right)+a^{7}+b^{7}}}\leq \dfrac{1}{2}$$

Generalization 1 Let $a,b,c$ be positive reals such that $abc=k$. Then prove that $$\sum_{cyc}{\dfrac{(ab)^{n}}{k\left(ab\right)^{n}+a^{2n+3}+b^{2n+3}}}\leq \dfrac{1}{k}$$

From rearrangement inequality, we have: $a^5+b^5\geq a^3b^2+a^2b^3 \implies a^5+b^5\geq a^2b^2(a+b)$ $$\implies \sum_{cyc} {\frac{ab}{a^5+b^5+ab}} \leq \sum_{cyc} {\frac{ab}{a^2b^2(a+b)+ab}}$$$$\implies \sum_{cyc} {\frac{ab}{a^5+b^5+ab}} \leq \sum_{cyc} {\frac{1}{ab(a+b)+1}}$$$$\implies \sum_{cyc} {\frac{ab}{a^5+b^5+ab}} \leq \sum_{cyc} {\frac{abc}{ab(a+b)+abc}}$$$$\implies \sum_{cyc} {\frac{ab}{a^5+b^5+ab}} \leq \sum_{cyc} {\frac{c}{a+b+c}}$$$$\implies \boxed{\sum_{cyc} {\frac{ab}{a^5+b^5+ab}} \leq 1}$$

Done

sqing wrote: Suppose that $a, b, c > 0$ such that $abc = 1$. Prove that $$ \frac{ab}{kab + a^5 + b^5} + \frac{bc}{kbc + b^5 + c^5} + \frac{ca}{kca + c^5 + a^5}\leq \frac{3}{k+2}$$Where $k\geq 1.$ Solution. Since $$x^5+y^5\ge x^2y^2(x+y)\,\,\forall x,y\in\mathbb{R}^+,$$using the assumption $abc=1$ we get that \begin{align*} & \frac{ab}{kab + a^5 + b^5} + \frac{bc}{kbc + b^5 + c^5} + \frac{ca}{kca + c^5 + a^5}\\ \le&\frac{ab}{kab + a^2 b^2(a+b)} + \frac{bc}{kbc + b^2c^2(b+c)} + \frac{ca}{kca + c^2a^(c+a)}\\ =&\frac{c}{kc + a+ b} + \frac{a}{ka + b+ c} + \frac{b}{kb + c+ a}. \end{align*}It remains to show that $$\frac{c}{kc + a+ b} + \frac{a}{ka + b+ c} + \frac{b}{kb + c+ a}\le\frac{3}{k+2},$$which is equivalent to $\left(\because k\ge1\right)$ $$3-\frac{(k-1)c}{kc + a+ b} - \frac{(k-1)a}{ka + b+ c} - \frac{(k-1)b}{kb + c+ a}\ge\frac{9}{k+2},$$namely, to show $$\frac{a+b+c}{kc + a+ b}+\frac{a+b+c}{ka + b+ c} +\frac{a+b+c}{kb + c+ a}\ge\frac{9}{k+2}.$$But the last inequality follows immediately from Cauchy-Schwartz’s inequality. $\blacksquare$

Notice that: $$\frac{4}{5}a^5 + \frac{1}{5} b^5 \ge a^4 b, \frac{1}{5}a^5 + \frac{4}{5} b^5 \ge a b^5$$Adding them gives $a^5+b^5 \ge a^4 b + ab^4$ (Essentially we use AM-GM, patching up Muir-head inequality ). Using this inequality, we have: $$\sum_{cyc} \frac{ab}{a^5+b^5+ab} \le \sum_{cyc} \frac{1}{a^3+b^3+1}$$ Notice that: $$\frac{2}{3} a^3 + \frac{1}{3} b^3 \ge a^2 b, \frac{1}{3} a^3 + \frac{2}{3} b^3 \ge a b^2$$Adding them gives: $a^3+b^3 \ge ab(a+b)$. Using this inequality, we have: $$\sum_{cyc} \frac{1}{a^3+b^3+1} = \sum_{cyc} \frac{abc}{a^3+b^3+abc} \le \sum_{cyc} \frac{abc}{ab(a+b+c)} = \sum_{cyc} \frac{a}{(a+b+c)} = 1.$$