<s>[INST] If $m+n=5$ and $mn=3$, find $\sqrt{\frac{n+1}{m+1}} + \sqrt{\frac{m+1}{n+1}}$? It is known that $m+n=5$ and $mn=3$. So what is the value of:

$$
\sqrt{\dfrac{n+1}{m+1}} + \sqrt{\dfrac{m+1}{n+1}}
$$

I think we're suppose to solve for the system of equations first, but I'm not getting any results that's useful.
 [/INST] Danny's solution is short and nice. Here is an alternate, longer and less elegant solution:
Let $\sqrt{\frac{n+1}{m+1}}+\sqrt{\frac{m+1}{n+1}}=t$. Squaring both sides, we have:
$$\frac{n+1}{m+1}+\frac{m+1}{n+1}+2=t^2$$
Simplifying, we have:
$$\frac{(n+1)^2+(m+1)^2}{(m+1)(n+1)}+2=t^2$$
i.e.,
$$\frac{n^2+2n+1+m^2+2m+1}{mn+m+n+1}+2=t^2$$
i.e., 
$$\frac{m^2+n^2+2(m+n)+2}{mn+m+n+1}+2=t^2$$
But, $m^2+n^2=(m+n)^2-2mn$. Therefore, we have:
$$\frac{(m+n)^2-2mn+2(m+n)+2}{mn+m+n+1}+2=t^2$$
Substituting for $m+n=5$ and $mn=3$, we have:
$$\frac{(5)^2-2(3)+2(5)+2}{3+5+1}+2=t^2$$
i.e., 
$$\frac{31}{9}+2=t^2$$
i.e.,
$$\frac{49}{9}=t^2$$
Thus, we get $t=\frac{7}{3}$. Note that we eliminate the spurious negative solution of $-\frac{7}{3}$ as $t$ must be greater than $0$.
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