Let $a,b,c\ge0: ab+bc+ca>0$. Prove that: $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{1}{a+b+c}+\frac{2}{\sqrt{ab+bc+ca}}$$
My attempt: Using C-S inequality:$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{9}{2(a+b+c)}$$ But $$\frac{9}{2(a+b+c)}\le \frac{2}{\sqrt{ab+bc+ca}}$$
Also, I try denote $f(a,b,c)=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}-\frac{1}{a+b+c}-\frac{2}{\sqrt{ab+bc+ca}}$ and set $f(0,b,c)=\frac{1}{b}+\frac{1}{c}-\frac{2}{\sqrt{bc}}$.
I want $f(a,b,c)\ge f(0,b,c)\ge0$. Here is my work: $f(a,b,c)- f(0,b,c)=\frac{-a}{a+b}+\frac{-a}{a+c}+2\left(\frac{\sqrt{ab+bc+ca}-\sqrt{bc}}{\sqrt{bc(ab+bc+ca)}}\right)-\frac{1}{a+b+c}$
But the rest seems complicated to me. I need your help to full my idea. Is there any better choice? Thanks for helping.
Best Answer
WLOG $ab+bc+ca=1$, then $$\Longleftrightarrow \dfrac{1}{a+b}+\dfrac{1}{a+b}+\dfrac{1}{c+a}-\dfrac{1}{a+b+c}\ge 2$$ $$\Longleftrightarrow \dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}-1\ge 2(a+b+c)$$ or $$2+\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}\ge 2(a+b+c)$$ since Use Cauchy-Schwarz inequality we have $$\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}\ge\dfrac{(a+b+c)^2}{2(ab+bc+ac)}=\dfrac{(a+b+c)^2}{2}$$ It suffices to show that: $$2+\dfrac{(a+b+c)^2}{2}\ge 2(a+b+c)$$ it is clear