Inequalities involving $L^p$ spaces.

I am reading "Minimax Methods in Critical Point Theory with Applications to Differential Equations" from Rabinowitz. On page 31 it says

"By standard inequalities, since $\mu>2$, for $\varepsilon>0$, $$\lVert u\rVert_{L^2(\Omega)}\leq a_8 \lVert u\rVert_{L^\mu(\Omega)}\leq a_9 K(\varepsilon)+\varepsilon \lVert u\rVert_{L^\mu(\Omega)}^\mu$$ where $K(\varepsilon)\to\infty$ as $\varepsilon\to0$."

The first inequality follows easily from Hölder inequality, however I have never heard of the second. How do you proof the second inequality?

My guess is this is Young's inequality, $$ab \leq \frac{a^p}{p}+\frac{b^q}{q}$$ ($1/p+1/q=1$, $a,b\geq 0$). If we choose $b=(\mu \varepsilon)^{1/\mu}\lVert u \rVert_{\mu}$, $q=\mu$, $a=a_8(\mu \varepsilon)^{-1/\mu}$, this turns into $$a_8 \lVert u \rVert_{\mu} \leq \frac{\mu-1}{\mu} \left(a_8(\mu \varepsilon)^{-1/\mu}\right)^{\mu/(\mu-1)} + \varepsilon \lVert u \rVert_{\mu}^{\mu};$$ in particular, the power of $\varepsilon$ is $-1/(\mu-1)<0$.
It suffices to show that for all $\varepsilon > 0,$ there is $K(\epsilon) > 0$ such that for all $a>0,$ we have $a \leq K(\varepsilon) + \varepsilon a^{\mu}.$ To show this, note that $\varepsilon a^{\mu} - a \rightarrow \infty$ as $a \rightarrow \infty$ (since $\mu > 2$), so there is $K(\epsilon)>0$ such that $\epsilon a^{\mu} > a$ whenever $a > K(\epsilon).$ Note this gives the desired $K(\epsilon).$
• @Chappers Whoops, I actually meant $\mu > 2$ since that was the assumption we had. Thanks for the correction. – ktoi Jun 12 '17 at 17:17