1
$\begingroup$

Given $a,b,c,d,x,y,z,w \geq 0$. By the Cauchy - Schwarz inequality we have that: \begin{equation} \label{1}\tag{1} \left( ax + by + cz + dw \right) ^{2} \leq \left( a^{2} + b^{2} + c^{2} + d^{2} \right) \left( x^{2} + y^{2} + z^{2} + w^{2} \right) . \end{equation} Also, by the Cauchy - Schwarz inequality we also obtain: \begin{equation} \label{2}\tag{2} \left( ax + by + cz + dw \right) ^{2} \leq 4 a^{2} x^{2} + 4 b^{2} y^{2} + 4 c^{2} z^{2} + 4 d^{2} w^{2} . \end{equation} In general it is better to use \eqref{1} or \eqref{2} as I am looking for the tightest estimate as possible. That is, generally we have \begin{align} \label{3}\tag{3} \left( ax + by + cz + dw \right) ^{2} & \leq \left( a^{2} + b^{2} + c^{2} + d^{2} \right) \left( x^{2} + y^{2} + z^{2} + w^{2} \right) \\ & \leq 4 a^{2} x^{2} + 4 b^{2} y^{2} + 4 c^{2} z^{2} + 4 d^{2} w^{2} \end{align} or \begin{align} \label{4}\tag{4} \left( ax + by + cz + dw \right) ^{2} & \leq 4 a^{2} x^{2} + 4 b^{2} y^{2} + 4 c^{2} z^{2} + 4 d^{2} w^{2} \\ & \leq \left( a^{2} + b^{2} + c^{2} + d^{2} \right) \left( x^{2} + y^{2} + z^{2} + w^{2} \right) . \end{align}

Edit: I forgot one assumption that $d = kc$ for some $k > 0$ and also $x,y,z,w > 0$ so that there is no trivial answer. The question is when will \eqref{3} and \eqref{4} corresponding to the choice of $k$.

$\endgroup$
2
  • $\begingroup$ Recall that the Cauchy-Schwarz inequality comes with an addendum: "... with equality iff ... are linearly dependent" $\endgroup$ Sep 17 '17 at 10:06
  • $\begingroup$ thanks. however these number are nothing relevant except that $d = kc$ so I don't know how to apply the linearly dependence $\endgroup$
    – JKay
    Sep 19 '17 at 12:21
1
$\begingroup$

Both your inequalities are wrong. $$\left( a^{2} + b^{2} + c^{2} + d^{2} \right) \left( x^{2} + y^{2} + z^{2} + w^{2} \right) \leq 4 a^{2} x^{2} + 4 b^{2} y^{2} + 4 c^{2} z^{2} + 4 d^{2} w^{2} $$ is wrong for $a=y=c=w\rightarrow0^+$. $$\left( a^{2} + b^{2} + c^{2} + d^{2} \right) \left( x^{2} + y^{2} + z^{2} + w^{2} \right) \geq 4 a^{2} x^{2} + 4 b^{2} y^{2} + 4 c^{2} z^{2} + 4 d^{2} w^{2} $$ is wrong for $b=c=d=y=z=w\rightarrow0^+$.

$\endgroup$
2
  • $\begingroup$ thank you. is there any chance so that one of the above inequality holds? as I can not check that these sequence are the same $\endgroup$
    – JKay
    Sep 19 '17 at 12:22
  • 1
    $\begingroup$ @khonglagica I think it can not be true because there are counterexamples. $\endgroup$ Sep 19 '17 at 12:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.