Proof of an inequality of $L^p$ norms For a general measure space, we define : $\|f\|_p= \left(\int\vert f\vert^p du\right)^{1/p}$. 
Let $0 < a < b < c < \infty$ and prove the following:
$$
\|f\|_b \leqslant \max\{\|f\|_a, \|f\|_c\}.
$$
Any help is appreciated because I dont understand the solution underneath
 A: We have that 
$$
\frac{1}{c}<\frac{1}{b}<\frac{1}{a}.
$$
Therefore, there exists $\alpha\in (0,1)$ such that 
$$
\frac{1}{b}=\frac{\alpha}{c}+\frac{1-\alpha}{a}.
$$
We claim that 
$$
\Vert f\Vert_b\leqslant \Vert f\Vert_c^\alpha\Vert f\Vert_a^{1-\alpha} (\ast)
$$
from where it follows that 
$$
\Vert f\Vert_b\leqslant\max\{\Vert f\Vert_a, \Vert f\Vert_c\}
$$
because 
$$
\Vert f\Vert_b\leqslant \Vert f\Vert_c^\alpha\Vert f\Vert_a^{1-\alpha}\leqslant \max\{\Vert f\Vert_a,\Vert f\Vert_c\}^{\alpha}\max\{\Vert f\Vert_a,\Vert f\Vert_c\}^{1-\alpha}=\max\{\Vert f\Vert_a,\Vert f\Vert_c\}.
$$
Now, (*) can be proved easily using Holder's Inequality as follows
$$
\int |f|^b=\int |f|^{\alpha b}|f|^{(1-\alpha)b}\leqslant \left(\int|f|^{\alpha b\frac{c}{\alpha b}}\right)^{\frac{\alpha b}{c}}\left(\int|f|^{(1-\alpha)b\frac{a}{(1-\alpha) b}}\right)^{\frac{(1-\alpha)b}{a}}=\left(\int|f|^{c}\right)^{\frac{\alpha b}{c}}\left(\int|f|^{a}\right)^{\frac{(1-\alpha)b}{a}}=\Vert f\Vert_c^{\alpha b}\Vert f\Vert_a^{(1-\alpha) b},
$$
where we have used the fact that
$$
\frac{\alpha b}{c}+\frac{(1-\alpha)b}{a}=1.
$$
Simiplifying we have the result.
