So here is the Question :-
How to prove that $$a + b + c + √3 \geq 8abc(\frac{1}{a^2 + 1} + \frac{1}{b^2+1} + \frac{1}{c^2 + 1})$$ where $a,b,c\geq 0$ are such that$ab + bc + ca \leq 1$ ?
What I tried :- I took the $$(\frac{1}{a^2 + 1} + \frac{1}{b^2+1} + \frac{1}{c^2 + 1})$$ part to the left hand side , and got $$(a + b + c + √3)(a^2 + b^2 + c^2 + 3) \geq 8abc$$
I already knew that $(a + b)(b + c)(c + a) \geq 8abc$ (which can be proved by AM-GM inequality) , so if i can able to show that :-
$$(a + b + c + √3)(a^2 + b^2 + c^2 + 3) \geq (a + b)(b + c)(c + a) \geq 8abc$$ , I am done . But now I don't know how to proceed further , or I am not sure whether $$(a + b + c + √3)(a^2 + b^2 + c^2 + 3) \geq (a + b)(b + c)(c + a)$$ is really true or not , in that case the proof won't work . Also I did not even use the fact that $ab + bc + ca \leq 1$ .
Can anyone help me on this ? Hints or Suggestions will be greatly appreciated ! .
Edit 01:- As @Muhammad Zuhair Khan mentioned out, transferring $(\frac{1}{a^2 + 1} + \frac{1}{b^2+1} + \frac{1}{c^2 + 1})$ to the LHS side does not give $(a + b + c + √3)(a^2 + b^2+ c^2 + 3)$ . So my approach was wrong only .
It's wrong. Try $a=b=c=1$.
We obtain: $$3+\sqrt3\geq12,$$ which is wrong.
But for $ab+ac+bc\leq1$ it's true.
If so, we need to prove that: $$a+b+c+\sqrt{3(ab+ac+bc)}\geq8abc\sum_{cyc}\frac{1}{(a+b)(a+c)}$$ or $$(a+b+c)\prod_{cyc}(a+b)+\sqrt{3(ab+ac+bc)}\prod_{cyc}(a+b)\geq16abc(a+b+c),$$ which is true because by AM-GM $$\prod_{cyc}(a+b)\geq8abc$$ and since $$\prod_{cyc}(a+b)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)$$ it's just $$\sum_{cyc}c(a-b)^2\geq0,$$ by AM-GM again we obtain: $$\sqrt{3(ab+ac+bc)}\prod_{cyc}(a+b)\geq\frac{8}{9}(a+b+c)\sqrt{3(ab+ac+bc)^3}\geq8abc(a+b+c).$$