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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 $k$.

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  • $\begingroup$ Recall that the Cauchy-Schwarz inequality comes with an addendum: "... with equality iff ... are linearly dependent" $\endgroup$ Sep 17, 2017 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, 2017 at 12:21

1 Answer 1

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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 $b=c=d=y=z=w\rightarrow0^+$.

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  • $\begingroup$ 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, 2017 at 12:22
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    $\begingroup$ @khonglagica I think it can not be true because there are counterexamples. $\endgroup$ Sep 19, 2017 at 12:23

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