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