<s>[INST] Minimum value of $(a+1)(b+1)$ 
If $a,b$ are the roots of the equation $$(\arcsin x+\arctan x)p^2+25p\pi+2(\arccos x+\operatorname{arccot} x)=0$$
Then minimum of $(a+1)(b+1)$

Try: $$a+b=\frac{25\pi}{\arcsin x+\arctan x}$$ and $$ab=\frac{2(\arccos x+\operatorname{arccot} x)}{\arcsin x+\arctan x}$$
So $$ab+a+b+1=\frac{25\pi+2(\arccos x+\operatorname{arccot} x)}{\arcsin x+\arctan x}$$
Could some help me to solve it, thanks
 [/INST] $x\neq0$, otherwise our equation has unique root.
Also,  $$\left(25\pi\right)^2-8(\arcsin{x}+\operatorname{arctan}x)(\arccos{x}+\operatorname{arccot}x)\geq$$
$$\geq\left(25\pi\right)^2-8\left(\frac{\arcsin{x}+\operatorname{arctan}x+\arccos{x}+\operatorname{arccot}x}{2}\right)^2=625\pi^2-2\pi^2>0,$$
which says that our equation has two real roots for all $x\neq0$ and $-1\leq x\leq 1.$
Now, $$(a+1)(b+1)=\frac{-25\pi+2(\arccos x+\operatorname{arccot} x)}{\arcsin x+\arctan x}+2-1=$$
$$=\frac{-25\pi+2(\arcsin{x}+\arccos x+\operatorname{arctan} x+\operatorname{arccot} x)}{\arcsin x+\arctan x}-1=$$
$$=\frac{-25\pi+2\left(\frac{\pi}{2}+\frac{\pi}{2}\right)}{\arcsin x+\arctan x}-1\rightarrow-\infty$$ for $x\rightarrow0^+$, which says that the minimum does not exist.
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