If $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\le 1$, prove that $(1+a^2)(1+b^2)(1+c^2)\ge 125$. 
QUESTION: Let $a,b,c$  be positive real numbers such that
$$\cfrac{1}{1+a}+\cfrac{1}{1+b}+\cfrac{1}{1+c}\le 1$$
Prove that $$(1+a^2)(1+b^2)(1+c^2)\ge 125$$ When does equality hold?


MY APPROACH: Firstly, let's try to squeeze out all the information we can from what is given. $$\frac{(1+b)(1+c)+(1+a)(1+c)+(1+a)(1+b)}{(1+a)(1+b)(1+c)}≤1$$ Multiplying this out, we get $$3+2(a+b+c)+(ab+bc+ca)≤1+(a+b+c)+(ab+bc+ca)+abc$$ $$\implies 2+(a+b+c)≤abc$$ Also, since $$1≥\sum_{cyc}\frac{1}{1+a}$$ Therefore by AM-GM,
$$1≥\sum_{cyc}\frac{3}{\sqrt[3]{(1+a)(1+b)(1+c)}}$$
$$\implies (1+a)(1+b)(1+c)≥27$$
That's all I ended up in.. At first, I thought Hölder's inequality could be employed, but that too requires the sum of the powers to be $=1$.. and that is not going to be useful in $(1+a^2)(1+b^2)(1+c^2)$ , since here the sum of powers add up to $3$..
I don't know what to do next.. Any help will be much appreciated..
 A: As $$abc\geq a+b+c+2\overset{AM-GM}{\geq} 4\sqrt[4]{2abc}\implies abc\geq 8$$Now,$$\prod_{cyc} (1+a^2)=\prod_{cyc} \left(1+\frac{a^2} 4+\frac{a^2} 4+\frac{a^2} 4+\frac{a^2} 4\right)\overset{AM-GM}{\geq}\prod_{cyc} 5\times \left(\frac{a}{2}\right)^{8/5}=125\times(abc/8)^{8/5}\geq 125$$
A: Another way.
Since by AM-GM  $$\prod_{cyc}\frac{a}{1+a}=\prod_{cyc}\left(1-\frac{1}{1+a}\right)\geq\prod_{cyc}\left(\frac{1}{1+b}+\frac{1}{1+c}\right)\geq$$
$$\geq\prod_{cyc}\frac{2}{\sqrt{(1+b)(1+c)}}=\frac{8}{\prod\limits_{cyc}(1+a)},$$ we see that $$abc\geq8$$ and by  Holder we obtain:
$$\prod_{cyc}(a^2+1)\geq\left(\sqrt[3]{a^2b^2c^2}+1\right)^3\geq(4+1)^3=125.$$
A: We need to prove that:
$$\sum_{cyc}(\ln(a^2+1)-\ln5)\geq0,$$ for which it's enough to prove that
$$\sum_{cyc}\left(\ln(a^2+1)-\ln5-\frac{36}{5}\left(\frac{1}{3}-\frac{1}{a+1}\right)\right)\geq0,$$ which is true because for any $a>0$ easy to show that:
$$\ln(a^2+1)-\ln5-\frac{36}{5}\left(\frac{1}{3}-\frac{1}{a+1}\right)\geq0.$$
Indeed, $$\left(\ln(a^2+1)-\ln5-\frac{36}{5}\left(\frac{1}{3}-\frac{1}{a+1}\right)\right)'=\frac{2(a-2)(5a^2+2a+9)}{5(a^2+1)(a+1)^2},$$ which gives $a_{min}=2$ and we are done!
I got the coefficient $-\frac{36}{5}$ by the following way.
Let $f(x)=\ln(a^2+1)-\ln5+\lambda\left(\frac{1}{3}-\frac{1}{a+1}\right).$
We know that $f(2)=0$ and we'll choose a value of $\lambda$ such that $2$ would be a double point,
for which we need $f'(2)=0,$ which gives $\lambda=-\frac{36}{5}.$
