For $a,b,c>0$. Find the minimize value of $$A=\frac{1}{\sqrt{a^2+b^2+c^2+1}}-\frac{2}{(a+1)(b+1)(c+1)}$$
By Cauchy-Schwarz and AM-GM we have:
$a+b+c+1\leq \sqrt{4(a^2+b^2+c^2+1)}\Rightarrow \frac{1}{\sqrt{a^2+b^2+c^2+1}}\leq \frac{2}{a+b+c+1}$
$(a+1)(b+1)(c+1) \leq \dfrac{(a+b+c+3)^3}{27}$
$\Rightarrow A\ge\frac{2}{a+b+c+1}+\frac{2}{\frac{\left(a+b+c+3\right)^3}{27}}$
I can't countinue. Help, thanks
Let $a=b=c\rightarrow0^+$.
Hence, $A\rightarrow-1$.
We'll prove that $$\frac{1}{\sqrt{a^2+b^2+c^2+1}}-\frac{2}{(a+1)(b+1)(c+1)}\geq-1.$$ Indeed, by AM-GM $$\frac{1}{\sqrt{a^2+b^2+c^2+1}}+1\geq\frac{2}{\sqrt[4]{a^2+b^2+c^2+1}}.$$ Thus, it's enough to prove that $$\frac{2}{\sqrt[4]{a^2+b^2+c^2+1}}\geq\frac{2}{(a+1)(b+1)(c+1)}$$ or $$(a+1)^4(b+1)^4(c+1)^4\geq a^2+b^2+c^2+1,$$ which is true because $$(a+1)^4(b+1)^4(c+1)^4\geq(a+1)^2(b+1)^2(c+1)^2\geq(a+b+c+1)^2\geq a^2+b^2+c^2+1.$$ The equality does not occur, which says that the minimum does not exist.
I think to find a maximal value is more interesting.
Let $a=b=c=1$.
Hence, $A=\frac{1}{4}$.
We'll prove that it's a maximal value.
Indeed, let $a+b+c=3u$.
Hence, by C-S, AM-GM and C-S we obtain: $$\frac{1}{\sqrt{a^2+b^2+c^2+1}}-\frac{2}{(a+1)(b+1)(c+1)}\leq$$ $$\leq\frac{1}{\sqrt{(\frac{(a+b+c)^2}{3}+1}}-\frac{2}{\left(\frac{a+b+c+3}{3}\right)^3}=$$ $$=\frac{1}{\sqrt{3u^2+1}}-\frac{2}{(u+1)^3}=\frac{2}{\sqrt{(3+1)(3u^2+1)}}-\frac{2}{(u+1)^3}\leq$$ $$\leq\frac{2}{3u+1}-\frac{2}{(u+1)^3}.$$ Thus, it remains to prove that: $$\frac{2}{3u+1}-\frac{2}{(u+1)^3}\leq\frac{1}{4},$$ which is $$(u-1)^2(3u^2+8u+1)\geq0.$$ Done!