Approximation for square roots of the form $\sqrt{a}+\sqrt{b}\approx \sqrt{c}$

Question

Just for fun, I tried to find $a, b, c \in \mathbb N$ such that $\sqrt{a}+\sqrt{b}\ne \sqrt{c}$ and $\sqrt{a}+\sqrt{b}$ approximates $\sqrt{c}$ as best as possible. In the trivial case $b=a+1, c=4a+2$ we have: $$ \big|\sqrt{a}+\sqrt{b}-\sqrt{c}\big|\approx\dfrac1{2c^{3/2}}. $$ There are many nontrivial solutions with the same approximation.

Q1. Is it true that for any $a, b, c\in \mathbb N$ such that $\sqrt{a}+\sqrt{b}\ne \sqrt{c}$ the following holds: $$ \big|\sqrt{a}+\sqrt{b}-\sqrt{c}\big|>\dfrac1{2c^{3/2}}? $$

I tried to look at cube roots but couldn't come up with any hypothesis. In this case, it is occasionally possible to find good approximations even for large numbers, for example: $$ \big|\sqrt[3]{11825}+\sqrt[3]{26471}-\sqrt[3]{145409}\big|<\dfrac1{\sqrt[3]{145409^7}}. $$

Q2. Is anything known about such approximation for cube or higher roots?

Answer 1

$\sqrt a+\sqrt b-\sqrt c$ is surely an algebraic number, and hence a root of a certain polynomial with integer coefficients, namely: $$x^8 - D_6x^6 + D_4x^4 - D_2 x^2 + D_0 = 0\text{, where}\\ \begin{aligned} D_6 &= 4(a+b+c);\\ D_4 &= 6(a^2 + b^2 + c^2) + 4(ab + bc + ac);\\ D_2 &= 4 (a^3 + b^3 + c^3 - a^2 b - b^2 a - a^2 c - c^2 a - b^2 c - c^2 b + 10 a b c). \end{aligned} $$ The last term is so ugly that I won't write it here. Also, it doesn't matter; what suffices to know is that it is an integer, and can't be $0$ (otherwise $0$ would be a root, which it isn't), so it is at least $1$. In fact, it is exactly $1$ for all triples $\{a, a+1, 4a+2\}$ and maybe for all sporadic good triples too.

Now, we may notice that at $|x|<1$ the positive term $+D_4x^4$ dwarfs the negative term $-D_6x^6$, hence the whole polynomial is bounded from below with $D_0 - D_2 x^2$. Therefore its roots must be no closer to zero than $\sqrt{\dfrac{D_0}{D_2}}$, or (considering what I said above of $D_0$'s ugliness, unimportance, and often being equal to $1$) simply $\dfrac1{\sqrt D_2}$.

Here my half-solution ends. I would very much like to say that the marginally ugly expression for $D_2$ (see above), which certainly has the right order of magnitude and even contains $4$ and $c^3$, is always greater than $4c^3$, which would give us the bound we were after. Pity I know this not to be true.

Answer 2

For the main question, $c^{-3/2}$ is certainly the right order of magnitude. Note that if $\sqrt a+\sqrt b-\sqrt c$ is small, then $\sqrt a+\sqrt b\approx\sqrt c$ and thus $\sqrt a+\sqrt b+\sqrt c\approx 2\sqrt c$. Therefore $$ a+b+2\sqrt{ab}-c = (\sqrt a+\sqrt b-\sqrt c)(\sqrt a+\sqrt b+\sqrt c) \approx 2\sqrt c(\sqrt a+\sqrt b-\sqrt c), $$ which reduces the question to the much simpler one of approximating non-integers $2\sqrt m$ (with $m$ in place of $ab$) by integers $n$ (with $n$ in place of $a+b-c$).

The best we can do is to have $n=2d+1$ odd and $m=d^2+d$, so that $n-2\sqrt m \sim \frac1{4d}$; taking $(a,b,c)=(d,d+1,4d+2)$ as in the OP yields $a+b+2\sqrt{ab}-c \sim -\frac1{4d} \sim -\frac1c$ and so $\sqrt a+\sqrt b-\sqrt c \sim -\frac1{2c^{3/2}}$.

While this answer hasn't done so, I believe this can be turned into a proof that the conjectured strict inequality is indeed satisfied.