Prove that $\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1$

Question

Hello ladies and gentlemen, here I have another inequality that I am struggling with:

Let $a,b,c>0$ Then $$\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1.$$

I try to show $$\frac{ab}{ab+b^2+ca}\le\frac{a}{a+b+c}$$ but this is doesn't work because $$\frac{ab}{ab+b^2+ca}=\frac{a}{a+b+\frac{ca}b}$$ which is in general not less than $\frac{a}{a+b+c}$.

So what can I do? Do I have to use some Hölder?

Answer 1

Multiply both sides with $\prod_{\text{cyc}} ab+b^2+ac$, subtract $ab\cdot(bc+c^2+ab)\cdot(ac+a^2+bc)$ from both sides and you get by expanding that the original inequality is equivalent to: \begin{equation}\tag 1\label 1\sum_{\text{cyc}} a^4bc+\sum_{\text{cyc}} a^3b^3\geq 2\sum_{\text{cyc}} a^3b^2c.\end{equation}

By AM-GM, we have \begin{align}2a^3b^3 + a^3c^3&\geq 3a^3b^2c, \\ 2a^3c^3 + b^3c^3&\geq 3a^2bc^3, \\ 2b^3c^3 + a^3b^3&\geq 3ab^3c^2.\end{align}

By summing these inequalities we get $$\sum_{\text{cyc}} a^3b^3\geq \sum_{\text{cyc}} a^3b^2c.$$ Similarly, we have $$2a^4bc+ab^4c\geq3a^3b^2c,\\\vdots$$

which gives $$\sum_{\text{cyc}} a^4bc\geq \sum_{\text{cyc}} a^3b^2c.$$ Hence, \eqref{1} is proven and hence also the original inequality.

Answer 2

Let $x = \frac ab, y = \frac bc, z = \frac ca$ and then the inequality is equivalent to : $$\sum_{x,y,z}\dfrac{1}{1+x+y}\leq 1$$ and after simplification it is the same as: $$\prod_{x,y,z}(x+y)\geq 2\sum_{x,y,z}x + 2.$$ But since $xyz = 1,$ we have $x+y+z\geq 3$ and $xy+yz+zx\geq 3,$ therefore: $$\prod_{x,y,z}(x+y) = \sum x\sum xy - 1\geq 3\sum x - 1\geq 2\sum x+2.$$ This is a bit terse, so you can work out the details.

Answer 3

Let $a=x^3$, $b=y^3$ and $c=z^3$.

Thus, by Muirhead $$\sum_{cyc}\frac{ab}{ab+b^2+ca}=\sum_{cyc}\frac{x^3y^3}{x^3y^3+\left(y^2\right)^3+(zx)^3}\leq$$ $$\leq\sum_{cyc}\frac{x^3y^3}{x^3y^3+\left(y^2\right)^2\cdot(zx)+(zx)^2\cdot(y^2)}=\sum_{cyc}\frac{x^2y}{x^2y+y^2z+z^2x}=1.$$