Find the maximum of the expression

Question

Let $a,b,c$ be real positive numbers so that $abc=1$. Find the maximum value that the following expression can attain:

$$\frac{a}{a^8+1}+\frac{b}{b^8+1}+\frac{c}{c^8+1}$$

My try:

I first though on apply a variable change so that $a=\frac{x}{y}$, $b= \frac{y}{z}$ and $c=\frac{z}{x}$. The problem is that the problem became harder for me:

$$\sum_{cyc} \frac{xy^7}{x^8+x^7}$$

Then I though on applying Holder in the denominator of the first expression, so it would look like:

$$\sum_{cyc} \frac{a}{a^8+1} \leq \sum_{cyc} \frac{2^7a}{(a+1)^8}$$

After that, I tried applying $a+1 \geq 2\sqrt{a}$. But the expression wasn't correct anymore.

Answer 1

We'll prove that $$\frac{a}{a^8+1}\leq\frac{3(a^6+1)}{4(a^{12}+a^6+1)}.$$ Indeed, we need to prove that $$\frac{1}{a^4+\frac{1}{a^4}}\leq\frac{3\left(a^3+\frac{1}{a^3}\right)}{4\left(a^6+\frac{1}{a^6}+1\right)}.$$ Let $a+\frac{1}{a}=2t$.

Thus, by AM-GM $t\geq1$ and we need to prove that: $$\frac{1}{16t^4-16t^2+2}\leq\frac{3(8t^3-6t)}{4((8t^3-6t)^2-2+1)}$$ or $$(t-1)(96t^6+32t^5-136t^4-40t^3+44t^2+8t-1)\geq0,$$ which is true because $$96t^6+32t^5-136t^4-40t^3+44t^2+8t-1\geq$$ $$\geq96t^6+32t^5-136t^4-40t^3+44t^2+4t=$$ $$=4t(t-1)(24t^4+32t^3-2t^2-12t-1)\geq0.$$ Thus, it's enough to prove that $$\sum_{cyc}\frac{3(a^6+1)}{4(a^{12}+a^6+1)}\leq\frac{3}{2}$$ or $$\sum_{cyc}\frac{a^6+1}{a^{12}+a^6+1}\leq2$$ or $$\sum_{cyc}\left(\frac{a^6+1}{a^{12}+a^6+1}-1\right)\leq2-3$$ or $$\sum_{cyc}\frac{a^{12}}{a^{12}+a^6+1}\geq1.$$ Now, let $a^6=\frac{x}{y}$ and $b^6=\frac{y}{z}$, where $x$, $y$ and $z$ are positives.

Thus, $c^6=\frac{z}{x}$ and we need to prove that $$\sum_{cyc}\frac{x^2}{x^2+xy+y^2}\geq1.$$ Now, by C-S we obtain: $$\sum_{cyc}\frac{x^2}{x^2+xy+y^2}=\sum_{cyc}\frac{x^2(x+z)^2}{(x^2+xy+y^2)(x+z)^2}\geq\frac{\left(\sum\limits_{cyc}(x^2+xy)\right)^2}{\sum\limits_{cyc}(x^2+xy+y^2)(x+z)^2}.$$ Id est, it's enough to prove that $$\left(\sum\limits_{cyc}(x^2+xy)\right)^2\geq\sum\limits_{cyc}(x^2+xy+y^2)(x+z)^2$$ or $$\sum_{cyc}(x^3y-x^2yz)\geq0$$ or $$\sum_{cyc}(x^3y-2x^2yz+z^2xy)\geq0$$ or $$\sum_{cyc}xy(x-z)^2\geq0.$$ Done!