Let $$f(n)=3(n-3)(n-5)(n-1+\sqrt{n^2-14n+61})$$ $$g(n)=(n-4)(3n^2-19n+34+\sqrt{(3n^2-19n+34)^2-48(n-3)(n-4)^2})$$ For $n\ge 11$, I have to prove that $f(n)<g(n).$
My Try:
I tried breaking into two parts,namely to prove that $$3(n-3)(n-5)(n-1)<(n-4)(3n^2-19n+34)$$ and $$3(n-3)(n-5)\sqrt{n^2-14n+61}<(n-4)\sqrt{(3n^2-19n+34)^2-48(n-3)(n-4)^2}$$ but couldn't succeed since first inequality is not true.
Kindly help.
For $n = 11, \cdots, 15$, the values of $f(n)$ and $g(n)$ can be computed directly to compare them.
Now suppose $n \geqslant 16$. Because$$ n^2 -14n + 61 < \left(n - \frac{19}{3}\right)^2 \Longleftrightarrow \frac{4}{3}n > \frac{188}{9} \Longleftrightarrow n > \frac{47}{3}, $$ then$$ f(n) < 3(n - 3)(n - 5)\left(n - 1 + n - \frac{19}{3}\right) = 6n^3 - 70n^2 + 266n - 330. \tag{1} $$
Also,\begin{align*} &\mathrel{\phantom{=}} (3n^2 - 19n + 34)^2 - 48(n - 3)(n - 4)^2\\ &= \left(3n^2 - 27n + \frac{182}{3}\right)^2 + 64n - \frac{1984}{9} > \left(3n^2 - 27n + \frac{182}{3}\right)^2, \end{align*} then\begin{align*} g(n) &= (n - 4)\left(3n^2 - 19n + 34 + \sqrt{(3n^2 - 19n + 34)^2 - 48(n - 3)(n - 4)^2}\,\right)\\ &> (n - 4)\left(3n^2 - 19n + 34 + 3n^2 - 27n + \frac{182}{3}\right)\\ &= (n - 4)\left(6n^2 - 46n + \frac{284}{3}\right) = 6n^3 - 70n^2 + \frac{836}{3} n - \frac{1136}{3}. \tag{2} \end{align*}
Combining (1) and (2),$$ g(n) - f(n) > \frac{38}{3} n - \frac{146}{3} > 0. $$