Inequality with integer numbers Let $a,b$ be two positive integer numbers such that $a\sqrt{3}>b\sqrt{7}$. 
Prove that  $(a\sqrt{3}-b\sqrt{7})(a+b)>1$.
Attempt. From the assumption we have $3a^2-7b^2\geq 1$. We try to argue that $3a^2-7b^2\geq 3$ to get
$$
a\sqrt{3}-b\sqrt{7}=\frac{3a^2-7b^2}{a\sqrt{3}+b\sqrt{7}}\geq\frac{3}{a\sqrt{3}+b\sqrt{7}}>\frac{1}{a+b}.
$$

$3a^2=7b^2+1$ is impossible for $b$ by modulo $3$.
$3a^2-2=7b^2$ is impossible for $a$ by modulo $7$.
Thus, $3a^2-7b^2\geq3$ and we are done!
The equality in the last inequality occurs for $a=55$ and $b=36$, which gives
$$S\geq91(55\sqrt3-36\sqrt7)=1.43....$$
Since $(55,36)$ is a minimal solution of the Pell $3a^2-7b^2=3$, 
it seems that we  got a minimal value of $S$. I have no a proof for this statement. 
