I read from The Cauchy-Schwarz master class: an introduction to the art of mathematical inequalities Book by J Michael Steele.
Theorem
If $1 < p < \infty$ and if $C$ is a constant such that \begin{eqnarray} \sum_{k=1}^n a_k x_k \leq C \left \{\sum_{k=1}^n |x_k|^p \right \}^{1/p} \end{eqnarray} for all $x_k,$ $1 \leq k \leq n,$ then for $q = p/(p-1)$ one has the bound \begin{eqnarray} \label{target_bound} \left \{\sum_{k=1}^n |a_k|^q \right \}^{1/q} \leq C \end{eqnarray}
Proof
For each $1 \leq k \leq n$ we should choose $x_k$ such that $a_k x_k = |x_k|^p;$ in other words, we set $x_k = sign(a_k)|a_k|^{p/(p-1)}$ where $sign(a_k)$ is $1$ if $a_k \geq 0$ and it is $-1$ if $a_k < 0.$ With this choice the condition of the theorem becomes \begin{eqnarray} \nonumber \sum_{k=1}^n |a_k|^{p/(p-1)} \leq C \{\sum_{k=1}^n |a_k|^{p/(p-1)}\}^{1/p}. \end{eqnarray} We can assume without loss of generality that the sum on the right > is nonzero, so it is safe to divide by that sum. The relation $1/p + 1/q = 1$ then confirms that we have indeed proved our target bound.
when i calculate $\DeclareMathOperator{\sign}{sign}a_k x_k = a_k \sign(a_k)\left|a_k\right|^{p/(p-1)} = \left|a_k\right|\,\left|a_k\right|^{p/(p-1)} = |a_k|^{(2p-1)/(p-1)}$ and $\left|\sign(a_k)|a_k|^{p/(p-1)}\right|^p = |a_k|^{p^2/(p-1)}$.
So, how this is right? I think he should choose $x_k = \sign(a_k)|a_k|^{1/(p-1)}$.
