A proof of Hölder's inequality and trying to understand thisLet's state the Hölder's inequality in the following way: | ∑ k = 1 n x k y k | ≤ ( ∑ k = 1 n | x k | p ) 1 / p ( ∑ k = 1 n | y k | q ) 1 / q where 1 ≤ p &lt; ∞ and q is such that 1 / p + 1 / q = 1 .Note that this just becomes the Cauchy-Schwarz for p = 1 / 2 = q

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A proof of Hölder&#039;s inequality and trying to understand thisLet&#039;s state the Hölder&#039;s inequality in the following way:      |        ∑          k      =      1        n        x    k        y    k        |    ≤  (      ∑          k      =      1        n        |        x    k              |        p        )          1              /            p        (      ∑          k      =      1        n        |        y    k              |        q        )          1              /            q      where   1  ≤  p  &amp;lt;  ∞ and q is such that   1      /    p  +  1      /    q  =  1 .Note that this just becomes the Cauchy-Schwarz for   p  =  1      /    2  =  q

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Step 1      a    i   and       b    i   are just a very cunning change of variable nothing more.As you can see       b    i    =      |        y    i              |              1              /            (      1      −      θ      )      Next       a    i    &amp;gt;  0 and       b    i    &amp;gt;  0 so       a    i    θ        b    i          1      −      θ        &amp;gt;  0 and transforming      a    i    θ        b    i          1      −      θ        ≤  θ      a    i        A          θ      −      1            B          1      −      θ        +  (  1  −  θ  )      b    i        A    θ        B          −      θ      into      ∑    i    (      a    i    θ        b    i          1      −      θ        )  ≤      ∑    i    (  θ      a    i        A          θ      −      1            B          1      −      θ        +  (  1  −  θ  )      b    i        A    θ        B          −      θ        )is completely licit.The inequality is finally obtained noticing that      ∑    i    (  θ      a    i        A          θ      −      1            B          1      −      θ        +  (  1  −  θ  )      b    i        A    θ        B          −      θ        )  =      A    θ        B          1      −      θ      Proof is obtained noticing that       |        ∑          k      =      1        n        x    k        y    k        |    ≤      ∑          k      =      1        n        |        x    k        |        |        y    k        |

A proof of Hölder's inequality and trying to understand this Let's state the Hölder's inequality in

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