Let us abbreviate the vector $(x,y,z,t)$ as $\mathbb{x}$$\mathbf{x}$. Combining Hölder's inequality and Young's inequality, $$ |\mathbb{x}|_3^{3/2} \leq |\mathbb{x}|_4 |\mathbb{x}|_2^{1/2} \leq \frac{3}{4} |\mathbb{x}|_4^{4/3} + \frac{1}{4}|\mathbb{x}|_2^2. $$$$ |\mathbf{x}|_3^{3/2} \leq |\mathbf{x}|_4 |\mathbf{x}|_2^{1/2} \leq \frac{3}{4} |\mathbf{x}|_4^{4/3} + \frac{1}{4}|\mathbf{x}|_2^2. $$ We do not have equality in the first inequality, because the entries of $\mathbb{x}$$\mathbf{x}$ are positive. Therefore $$ |\mathbb{x}|_3^{3/2} < \frac{3}{4} |\mathbb{x}|_4^{4/3} + \frac{1}{4}|\mathbb{x}|_2^2. $$$$ |\mathbf{x}|_3^{3/2} < \frac{3}{4} |\mathbf{x}|_4^{4/3} + \frac{1}{4}|\mathbf{x}|_2^2. $$ Rearranging, we obtain the desired inequality. In this last step we use that the denominator is positive, which is another application of Hölder's inequality: $$ |\mathbb{x}|_3^{3/2} \leq |\mathbb{x}|_4^{4/3} |\mathbb{x}|_1^{1/6}, $$$$ |\mathbf{x}|_3^{3/2} \leq |\mathbf{x}|_4^{4/3} |\mathbf{x}|_1^{1/6}, $$ where we do not have equality as before, and $|\mathbb{x}|_1=1$$|\mathbf{x}|_1=1$ by assumption.