Symmetric Inequality in $\mathbb{R}$

Solution 1:

You can use the rearrangement Inequality. Note $$\frac{ab}{a^5+b^5+ab}\le \frac{ab}{a^4b+ab^4+ab}\\ \frac{bc}{b^5+c^5+bc}\le \frac{bc}{b^4c+bc^4+bc}\\ \frac{ac}{a^5+c^5+ac}\le \frac{ac}{a^4c+ac^4+ac}\\ $$ therefore $$\sum_{cyc}\frac{ab}{a^5+b^5+ab}\le \sum_{cyc}\frac{ab}{a^4b+ab^4+ab}=\sum_{cyc}\frac{1}{a^3+b^3+1}=\sum_{cyc}\frac{1}{a^3+b^3+\color{red}{abc}}\tag 1$$ Similarly $$\frac{1}{a^3+b^3+abc}\le \frac{1}{a^2b+ab^2+abc}\\ \frac{1}{b^3+c^3+abc}\le \frac{1}{b^2c+bc^2+abc}\\ \frac{1}{a^3+c^3+abc}\le \frac{1}{a^2c+ac^2+abc}\\ $$ thus $$\sum_{cyc}\frac{1}{a^3+b^3+abc}\le\sum_{cyc}\frac{1}{a^2b+ab^2+abc}\le \sum_{cyc}\frac{1}{ab}\left(\frac{1}{a+b+c}\right)\le\underbrace{\frac{1}{a+b+c}\sum_{cyc}\frac{1}{ab}}_{\frac{1}{a+b+c}\times\frac{a+b+c}{abc}=\frac{1}{abc}}=1\tag 2 $$ $(1)$ and $(2)$ $$\sum_{cyc}\frac{ab}{a^5+b^5+ab}\le1$$

Solution 2:

This is an IMO 1996 Shortlist problem. Here is a solution. Another one here.

Well if you consider the approach in the first link I shared, you can do everything with some bunch of AGMs. Consider the following

\begin{align*} 2a^4 + 2b^4&= a^4 + b^4 +a^4 + b^4\\ &\geq a^4 + b^4 + 2a^2b^2 \\ &= (a^4 + a^2b^2) + (b^4 + a^2b^2) \\ &\geq 2a^3b + 2ab^3 \\ \end{align*}

Here we showed $a^4 + b^4 - a^3b -ab^3 \geq 0$. Therefore \begin{align*} a^5 + b^5 &= (a+b)(a^4 + b^4 - a^3b -ab^3 + a^2b^2)\\ &\geq (a+b)a^2b^2 \end{align*} Which means \begin{align*} \frac{ab}{a^5 + ab +b^5} &\leq \frac{ab}{(a+b)a^2b^2 + ab}\\ &\leq \frac{1}{(a+b)ab + 1}\\ &= \frac{c}{a+b+c}\\ \end{align*} Note: You can do without "Rearrrengement".