Extremal problem with positive integer numbers Let $a,b$ be two positive integer numbers such that $a\sqrt{3}-b\sqrt{7}>0$. Find the minimum value of
$$
S=(a\sqrt{3}-b\sqrt{7})(a+b).
$$
Attempt I have tried and guess that the minimum value of $S$ is $(55+36)(55\sqrt{3}-36\sqrt{7})$, where $(55,36)$ is the integer solution of the Pell equation $3a^2-7b^2=3$.

We need to minimise
$$S = \sqrt3 \left(1+\frac{a}b\right)\cdot \color{blue}{b^2\left(\frac{a}b-\sqrt{\frac73} \right)}$$
There are infinitely many convergents from the regular continued fraction expansion of $\alpha = \sqrt{\frac73}$ which keep the terms in blue bounded, i.e. satisfy $b^2\left(\frac{a}b-\alpha \right) < \frac12$, and only convergents satisfy this.  Further as we take higher convergents the terms not in blue gets lower.  Hence it is enough to seek the minimum (or greatest lower bound) from these convergents.
We have the bounds for convergents $a_n/b_n$
$$\frac1{2b_n^2\alpha-1}> \left|\frac{a_n}{b_n} - \alpha\right| > \frac1{2b_n^2\alpha+1} $$
$$\implies \frac{1}{2\alpha -1/b_n^2} > b_n^2\left|\frac{a_n}{b_n} - \alpha\right| > \frac1{2\alpha+1/b_n^2}$$
$$\implies \frac{\sqrt3 (1+a_n/b_n)}{2\alpha-1/b_n^2} > S_n > \frac{\sqrt3 (1+a_n/b_n)}{2\alpha+1/b_n^2}$$
As we take better convergents, we get $a_n/b_n \to \alpha$ and $b_n \to \infty$, so the greatest lower bound is 
$$S_n \to S_{glb} = \frac{\sqrt3(1+\alpha)}{2\alpha} = \frac{\sqrt3}2 + \frac3{2\sqrt7}\approx 1.432972113$$ 
