# Calculate the maximum value of $\sum_{cyc}\frac{bc}{(b + c)^3(a^2 + 1)}$ where $a, b, c \in \mathbb R^+$ satisfying $abc = 1$.

Calculate the maximum value of $$\large \frac{bc}{(b + c)^3(a^2 + 1)} + \frac{ca}{(c + a)^3(b^2 + 1)} + \frac{ab}{(a + b)^3(c^2 + 1)}$$ where $$a, b, c$$ are positives satisfying $$abc = 1$$.

We have that $$\sum_{cyc}\frac{bc}{(b + c)^3(a^2 + 1)} \le \frac{1}{2} \cdot \sum_{cyc}\frac{1}{(b + c)(a^2 + 1)} \le \sum_{cyc}\frac{1}{(b + c)(a + 1)^2}$$

$$= \sum_{cyc}\frac{1}{a(ab + ca + b + c)(bc + 1)} \le \dfrac{1}{2} \cdot \sum_{cyc}\frac{1}{a\sqrt{bc}(ab + ca + 2\sqrt{bc})}$$

$$= \frac{1}{2} \cdot \sum_{cyc}\frac{1}{a\sqrt a(b + c) + 2}$$

That's all I got, not because I can't go for more, but since I went overboard with this, there's no use trying to push for more.

The inequality $$\sum_{cyc}\frac{x}{x^4+1}\leq\frac{3}{2}$$ for positives $$x$$, $$y$$ and $$z$$ we can prove also by the following way.

We'll prove that $$\frac{x}{x^4+1}\leq\frac{3(x^2+1)}{4(x^4+x^2+1)}$$ is true for all positive $$x$$.

Indeed, let $$x^2+1=2ux.$$

Thus, by AM-GM $$u\geq1$$ and it's enough to prove that $$3\cdot2u\cdot(4u^2-2)\geq4(4u^2-1)$$ or $$6u^3-4u^2-3u+1\geq0$$ or $$(u-1)(6u^2+2u-1)\geq0,$$ which is obvious.

Id est, it's enough to prove that $$\sum_{cyc}\frac{x^2+1}{x^4+x^2+1}\leq2$$ or $$\sum_{cyc}\frac{x+1}{x^2+x+1}\leq2$$ or $$\sum_{cyc}\left(\frac{x+1}{x^2+x+1}-1\right)\leq-1$$ or $$\sum_{cyc}\frac{x^2}{x^2+x+1}\geq1$$ or $$\sum_{cyc}(x^2y^2-x)\geq0$$ or $$\sum_{cyc}(x^2y^2-x^2yz)\geq0$$ or $$\sum_{cyc}z^2(x-y)^2\geq0$$ and we are done!

Let $$a=b=c=1$$.

Thus, we obtain a value $$\frac{3}{16}.$$

We'll prove that it's a maximal value.

Indeed, by AM-GM $$\sum_{cyc}\frac{bc}{(b+c)^3(a^2+1)}\leq\sum_{cyc}\frac{bc}{8\sqrt{b^3c^3}(a^2+1)}=\sum_{cyc}\frac{\sqrt{a}}{8(a^2+1)}.$$ Id est, it's enough to prove that $$\sum_{cyc}\frac{a}{a^4+1}\leq\frac{3}{2},$$ where $$a$$, $$b$$ and $$c$$ are positives such that $$abc=1$$.

We have $$\frac{3}{2}-\sum_{cyc}\frac{a}{a^4+1}=\sum_{cyc}\left(\frac{1}{2}-\frac{a}{a^4+1}-\frac{1}{2}\ln{a}\right).$$ Let $$f(x)=\frac{1}{2}-\frac{x}{x^4+1}-\frac{1}{2}\ln{x}.$$

Thus, easy to see that $$f(x)\geq0$$ for all $$0

and from here our inequality is proven for $$\max\{a,b,c\}<2$$.

Let $$a\geq2$$.

Thus, by AM-GM: $$\sum_{cyc}\frac{a}{a^4+1}\leq\frac{2}{2^4+1}+2\cdot\frac{x}{4\sqrt[4]{x^4\cdot\left(\frac{1}{3}\right)^3}}=\frac{2}{17}+\frac{\sqrt[4]{27}}{2}<\frac{3}{2}$$ and we are done!

• @Lê Thành Đạt It would be another problem. Fix please that our variables are positive real. – Michael Rozenberg Sep 14 '19 at 14:13
• I would still like to know whether replacing $a$ with $bc$ in the denominator would fix the problem. – Lê Thành Đạt Sep 14 '19 at 14:24
• @Lê Thành Đạ Is it means that we need to work with $\sum\limits_{cyc}\frac{a}{(b+c)^3(b^2c^2+1)}$? – Michael Rozenberg Sep 14 '19 at 14:52
• Wait, I meant the numerator, not the denominator. Sorry about that. – Lê Thành Đạt Sep 14 '19 at 16:31
• @Lê Thành Đạt A maximal value of $\sum\limits_{cyc}\frac{bc}{(b+c)^3(a^2+1)}$ is equal to $\frac{3}{16}.$ – Michael Rozenberg Sep 14 '19 at 17:44