Assume $f\in L^p(\mathbb{R}^n)\cap L^q(\mathbb{R}^n)$, $1\leq p<q\leq\infty$. Prove that for any $p<r<q$, there exists $0<\lambda<1$ such that $$\large\|f\|_r\leq\|f\|_p^\lambda\|f\|_q^{1-\lambda}$$
What I tried is to first consider the case $q<\infty$. Then there exists $0<\lambda<1$ such that $$r=\lambda p+(1-\lambda)q$$
Then \begin{align*} \|f\|_r^r&=\int|f|^{\lambda p+(1-\lambda)q}\\ &=\|f^{\lambda p}\cdot f^{(1-\lambda)q}\|_1\\ &\leq\|f^{\lambda p}\|_{1/\lambda}\|f^{(1-\lambda)q}\|_{1/(1-\lambda)}\tag{by Holder's Inequality}\\ &=\left(\int|f|^p\right)^\lambda\left(\int|f|^q\right)^{1-\lambda} \end{align*}
Thus $\large\|f\|_r\leq\left(\int|f|^p \right)^\frac{\lambda}{\lambda p+(1-\lambda)q}\left(\int|f|^q\right)^\frac{1-\lambda}{\lambda p+(1-\lambda)q}$ which is quite similar but yet not the same as what is required.
Thanks for any help.
Try reverse engineering: $$ \|f\|_p^λ·\|f\|_q^{1-λ}=\||f|^λ\|_{p/λ}·\||f|^{1-λ}\|_{q/(1-λ)} \ge\||f|^λ·|f|^{1-λ}\|_r $$ which is valid as $p/λ>p\ge 1$, $q/(1-λ)>q\ge 1$. The third number $r$ is connected to the other exponents by $$ \frac1r=\frac1{p/λ}+\frac1{q/(1-λ)}=\frac{λ}{p}+\frac{1-λ}{q} $$ by Young's inequality. This allows to easily compute $ λ$.