Prove using Cauchy-Schwarz inequality that if $\{x_i\}$, $\{y_i\}$, $\{z_i\}$, $\{u_i\}$ and $\{w_i\}$ are all positive reals, $i=1,2,\cdots n$ then show that


Attempt: Using Cauchy-Schwarz inequality on the sets $\{x_i\}$, $\{y_i.z_i.u_i.w_i\}$, we have, $$\left(\sum_{i=1}^nx_iy_iz_iu_iw_i\right)^2\leq\sum_{i=1}^nx_i^2.\sum_{i=1}^ny_i^2z_i^2.u_i^2.w_i^2=\sum_{i=1}^nx_i^2\sum_{i=1}^n\left(y_iz_i.u_i.w_i\right)^2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$$ But how to adjust the second term of the RHS of (1)?

Please help.

  • $\begingroup$ Which is larger: $\sum_{i=1}^n (y_i z_i w_i)^2$ or $\left(\sum_{i=1}^n (y_i z_i w_i)\right)^2$? keep in mind: positive reals $\endgroup$ – Prototank Feb 21 '16 at 6:59
  • $\begingroup$ @Prototank yes the 2nd one. But how to get the desired result. Would you suggest in more details please. $\endgroup$ – user1942348 Feb 21 '16 at 7:10
  • 2
    $\begingroup$ If the second is larger as you say, then this should also be true: $\sum_{i=1}^nx_i^2\sum_{i=1}^n\left(y_iz_i.u_i.w_i\right)^2\leq \sum_{i=1}^nx_i^2\left(\sum_{i=1}^ny_iz_i.u_i.w_i\right)^2$ In which case, apply the inequality again. $\endgroup$ – Prototank Feb 21 '16 at 7:18
  • $\begingroup$ @Prototank Ok I understand the step. Please explore "apply the inequality again"...please $\endgroup$ – user1942348 Feb 21 '16 at 9:18
  • $\begingroup$ Do as you already have done. View $y_i z_i u_i w_i$ as a product of two real numbers and apply the Cauchy Schwarz inequality to the rightmost term again. $\endgroup$ – Prototank Feb 21 '16 at 15:26

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