For $a,b,c>0$ and $a+b+c=1.$ Prove$:$ $$\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\geqq \frac{1}{ab+bc+ca}$$
This inequality is easy and there are two nice proof by AM-GM or by C-S also.
SOS also help here$:$
$$\text{LHS}-\text{RHS}=\frac{3\Big[\sum\limits_{cyc} (ab+bc-2ca)^2 + (ab+bc+ca) \sum\limits_{cyc} (a-b)^2 \Big]}{2 \prod (ab+2c^2 +2c)}+\frac{\prod (a-b)^2}{(ab+bc+ca) \prod (ab+2c^2 +2c)} \geqq 0$$
By SOS$,$ I can only found this stronger with same condition$:$ $$\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\geqq \frac{1}{ab+bc+ca}+\frac{\prod (a-b)^2}{(ab+bc+ca) \prod (ab+2c^2 +2c)}$$
But it's is very easy so I wanna to find another harder version for it$?$
Thanks for a real lot!
Best Answer
An improvement of your last inequality is the following: For $a, b, c > 0$ with $a+b+c=1$, \begin{align} &\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\\ \ge\ & \frac{1}{ab+bc+ca} + 4\cdot \frac{\prod (a-b)^2}{(ab+bc+ca) \prod (ab+2c^2 +2c)}.\tag{1} \end{align} $4$ is the best constant in the sense that we cannot replace it with a constant strictly larger than $4$. (1) is verified by Mathematica. One may find a step-by-step proof of it.
The following inequality is also true: For $a, b, c > 0$ with $a+b+c=1$, \begin{align} &\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\\ \ge\ & \frac{1}{ab+bc+ca} + \frac{27}{32}[(a-b)^2+(b-c)^2+(c-a)^2].\tag{2} \end{align} $\frac{27}{32}$ is the best constant in the sense that we cannot replace it with a constant strictly larger than $\frac{27}{32}$. (2) is verified by Mathematica. One may find a step-by-step proof of it.
In general, we first find a function $f(a, b, c)\ge 0$ with $f(1/3, 1/3, 1/3)=0$, then determine $$\alpha = \inf_{a, b, c > 0; \ a+b+c=1}\frac{\dfrac{1}{ab+2c^{2}+2c}+\dfrac{1}{bc+2a^{2}+2a}+\dfrac{1}{ca+2b^{2}+2b} - \dfrac{1}{ab+bc+ca}}{f(a,b,c)}.$$ Then the following inequality is true: For $a, b, c > 0$ with $a+b+c=1$, $$\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b} \ge \frac{1}{ab+bc+ca} + \alpha f(a,b,c).\tag{3}$$