I want to show the following inequality $$ \left\| \boldsymbol{v} \right\| _{2/3}^{2}\ge \left\| \boldsymbol{v} \right\| _{1}^{3} \tag 1 $$ where the norm is defined as $\left\| \boldsymbol{v} \right\| _p \equiv \left[ \sum_k{\left| v_k \right|^p} \right] ^{1/p}$ and $\boldsymbol{v}$ is a vector which can satisfy $\left\| \boldsymbol{v} \right\| _2=1$ if this condition is necessary to show the eq(1).
I've tried to use Hölder's inequality for Schatten norm but failed. I wonder if I missed something or should I use other inequality? But the Hölder's inequality is the only inequality I am familiar with for the Schatten norm.
Some more information about Hölder's inequality can be found in this post.
Actually we don't need such complex inequalities. Instead, we only need to prove it by definition. By definition, we have $$ \left\| \boldsymbol{v} \right\| _{2/3}^{2}=\left( \sum_k{\left| v_k \right|^{\frac{2}{3}}} \right) ^3 \\ \left\| \boldsymbol{v} \right\| _{1}^{3}=\left( \sum_k{\left| v_k \right|} \right) ^3 $$ Hence we only need to show $$ \sum_k{\left| v_k \right|^{\frac{2}{3}}}\ge \sum_k{\left| v_k \right|} $$ Using the fact that $\left\| \boldsymbol{v} \right\| _2=1$, we have $\left| v_k \right|\le 1,\forall k$ hence $\left| v_k \right|^{\frac{2}{3}}\ge \left| v_k \right|,\forall k$ which completes the proof.