Using a Inequality to generate a generalized inequality

Today, we are going to generalize this identity, which stipulates:

${\displaystyle \left(\frac{1}{b-a}\int_{a}^{b}f\left(x\right)^{t}dx\right)^{\frac{1}{t}}>\left(\frac{1}{b-a}\int_{a}^{b}f\left(x\right)^{u}dx\right)^{\frac{1}{u}}}$ if ${\displaystyle t>u}$.

Like always, we will start by doing this:

${\displaystyle \left(\frac{1}{b-a}\right)^{\frac{1}{t}}\left(\int_{a}^{b}f\left(x\right)^{t}dx\right)^{\frac{1}{t}}>\left(\frac{1}{b-a}\right)^{\frac{1}{u}}\left(\int_{a}^{b}f\left(x\right)^{u}dx\right)^{\frac{1}{u}}}$

Then we will use negative exponents, just temporarily:

${\displaystyle \left(b-a\right)^{-\frac{1}{t}}\left(\int_{a}^{b}f\left(x\right)^{t}dx\right)^{\frac{1}{t}}>\left(b-a\right)^{-\frac{1}{u}}\left(\int_{a}^{b}f\left(x\right)^{u}dx\right)^{\frac{1}{u}}}$

${\displaystyle \frac{\left(\int_{a}^{b}f\left(x\right)^{t}dx\right)^{\frac{1}{t}}}{\left(b-a\right)^{\frac{1}{t}}}>\frac{\left(\int_{a}^{b}f\left(x\right)^{u}dx\right)^{\frac{1}{u}}}{\left(b-a\right)^{\frac{1}{u}}}}$

Now finally, witness the magic that happens if we raise both sides to the power of ${\displaystyle tu}$:

${\displaystyle \frac{\left(\int_{a}^{b}f\left(x\right)^{t}dx\right)^{\frac{tu}{t}}}{\left(b-a\right)^{\frac{tu}{t}}}>\frac{\left(\int_{a}^{b}f\left(x\right)^{u}dx\right)^{\frac{tu}{u}}}{\left(b-a\right)^{\frac{tu}{u}}}}$

${\displaystyle \frac{\left(\int_{a}^{b}f\left(x\right)^{t}dx\right)^{u}}{\left(b-a\right)^{u}}>\frac{\left(\int_{a}^{b}f\left(x\right)^{u}dx\right)^{t}}{\left(b-a\right)^{t}}}$

${\displaystyle \left(b-a\right)^{t}\left(\int_{a}^{b}f\left(x\right)^{t}dx\right)^{u}>\left(b-a\right)^{u}\left(\int_{a}^{b}f\left(x\right)^{u}dx\right)^{t}}$

Special Case: ${\displaystyle b=a+1}$

When this happens, the inequality becomes:

${\displaystyle \left(\int_{a}^{a+1}f\left(x\right)^{t}dx\right)^{u}>\left(\int_{a}^{a+1}f\left(x\right)^{u}dx\right)^{t}}$, for ${\displaystyle t>u}$

This special case gives a whole slew of identities. However, we will not give them in this blog post; that is for you to figure out!