For an integer $n$, denote by $P^+(n)$ the largest prime divisor of $n$. Then we have the following:

There exists some $c>0$, such that for all $x$ sufficiently large the number of integers $n\in[x, x+0.1\sqrt{x}]$ with $P^+(n)>x^{1/2+c}$ is $\geq c\sqrt{x}$.

The proof is quite standard: You essentially need a non-trivial bound for $$ \sum_{n\leq x^{1/2+c}}\Lambda(n)\left(\left[\frac{x+0.1\sqrt{x}}{n}\right]-\left[\frac{x}{n}\right]-\frac{0.1\sqrt{x}}{n}\right). $$ Now express $\Lambda$ by Vaughan's identity, approximate $[\cdot]$ by exponential sums, and bound the latter by Weyl-van der Corput.

However, while conceptually simple, the computations take at least two pages, which I would rather avoid since

a) journal space is precious

b) the computations are likely to distract the reader from the more algebraic main arguments,

c) the question is so natural that someone probably spent a lot of time on some quantitative version of it.

So my question is: Can anyone give me a reference for such a result? Note that for the application the value of $c$ is of no importance, while 0.1 should not be replaced by something much larger.

Thank you in advance!