Disproving using the definition of O Formally show that $0.1n + 10\sqrt{n}$ is not $O(\sqrt{n})$ using the definition of $O$ only.

I cannot find much on how to solve a problem like this, nor do i know how. Am I supposed to show some contradiction by trying to prove it is $O(\sqrt{n})$? Any help on how to answer this would be appreciated. 

This is about measures of complexity, so we're interested in asymptotic behavior as $n \to +\infty$, and we assume that all the functions take only non-negative values ($\forall n, f(n) \ge 0$).
Here is one possible definition of big O (your definition might vary in details):


  Let $f(n),g(n)$ be two real-valued functions such that $f(n),g(n) &gt; 0$ for all $n$. We say that $f(n) = O(g(n))$ if there exists an integer $N$ and a real $C &gt; 0$ such that for all $n \geq N$, $f(n) \leq Cg(n)$.


Using the rules of logic, we get a definition of the negation of big O:


  Let $f(n),g(n)$ be two real-valued functions such that $f(n),g(n) &gt; 0$ for all $n$. We say that $f(n)$ is not big $O(g(n))$ if for every integer $N$ and real $C &gt; 0$ there exists $n \geq N$ such that $f(n) &gt; Cg(n)$.


This is what you should prove. For each integer $N$ and real $C &gt; 0$ you should find $n \geq N$ such that $0.1n + 10\sqrt{n} &gt; C\sqrt{n}$.