Prove the inequality $0<(m+n)/(mn-1)\le 3$ for $m,n\in\mathbb N $ with $mn\ne 1$ Prove the simple inequality:  $0<\frac{n_1+n_2}{n_1n_2-1}\leq 3$ for all $n_1,n_2\in\mathbb{N}$ with $n_1n_2\neq1$
Progress
I have checked the inequality for $n_1,n_2=1,2,3$
 A: Set: $$n_3=\max\{n_1,n_2\}$$
So
$$n_1+n_2\le2n_3\,\,\,\text{and}\,\,\,n_1n_2\ge n_3$$
Therefore, for $n_3\ge3$:
$$\frac{n_1+n_2}{n_1n_2 - 1}\le\frac{2n_3}{n_3 - 1}\le3$$
And as you noted, you already checked that the desired inequality is correct when $n_3<3$.
A: Let $A=\frac{n_1+n_2}{n_1n_2-1}$ with $n_1,n_2\in\mathbb{N}$ and $n_1n_2>1$. Then $A>0$ because both its numerator and denominator are strictly positive. Moreover,
$$
A\leq 3\iff 3n_1n_2-3\geq n_1+n_2\iff n_1(n_2-1)+n_2(n_1-1)+n_1n_2-3\geq 0.
$$
The last inequality is true because the nonnegative terms $n_1(n_2-1)$ and $n_2(n_1-1)$ cannot both be $0$ (otherwise we will have $n_1=n_2=1$.) And so
$$
n_1(n_2-1)+n_2(n_1-1)+n_1n_2-3\geq 1+n_1n_2-3\geq 1+2-3=0.
$$
The last inequality above uses $n_1n_2>1$ which implies $n_1n_2\geq 2$.
A: Use mathematical induction.  Hold $n_2\ge2$ fixed.  Then, for $n_1=1$ and $n_1=2$ the inequality holds.  Assume the inequality holds for $n_1=m$.  Now, investigate $$\frac{m+1+n_2}{(m+1)n_2-1}=\frac{(m+n_2)+1}{(mn_2-1)+n_2}$$But since $1<n_2$, then $$\frac{(m+n_2)+1}{(mn_2-1)+n_2}\le\frac{3(mn_2-1)+1}{(mn_2-1)+n_2}<3$$  This works for any $n_2>1$.  The case for $n_2=1$ follows by interchanging $n_2$ and $n_1$. QED
A: As all numbers are positive, you need to prove
$$n_1+n_2+3\leq 3n_1n_2$$
You already covered the case when $n_1n_2=2$. 
For the other cases you have
$$n_1 \leq n_1n_2$$
$$n_2 \leq n_1n_2$$
$$3 \leq n_1n_2$$
Add them together.
