<s>[INST] 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..
 [/INST] 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:
$$\prod_{cyc}(a^2+1)\geq\left(\sqrt[3]{a^2b^2c^2}+1\right)^3\geq(4+1)^3=125.$$
</s>