Possible mistake in the solution of Baby Rudin Ch. 6 Ex. 11

Question

This is exercise 11, Chapter 6 in Baby Rudin:

Let $alpha$ be a fixed increasing function on $[a, b]$. For $u in mathscr{R}(alpha)$, define
$$ lVert u rVert_2 = left{ int_a^b lvert u rvert^2  mathrm{d} alpha right}^{1/2}. $$
Suppose $f, g, h in mathscr{R}(alpha)$, and prove the triangle inequality
$$ lVert f-h rVert_2 leq  lVert f-g rVert_2 + lVert g-h rVert_2 $$
as a consequence of the Schwarz inequality

where by "Schwarz inequality," Rudin means the following version of Holder's inequality:

$$leftlvert int_a^b f g  mathrm{d} alpha rightrvert leq left{ int_a^b lvert f rvert^2  mathrm{d} alpha right}^{1/2} left{ int_a^b lvert g rvert^2  mathrm{d} alpha right}^{1/2}. $$

The solution to this question in the solutions manual is:

In the first display on the left-hand side, it should read $||f-h||_2^2$ instead of $||f-h||_2^1$. My question is about the only inequality in the proof. In particular, how is it true that $$int_a^b |f-g||g-h|  dalpha le left{ int_a^b lvert f-g rvert^2  mathrm{d} alpha right}^{1/2} left{ int_a^b lvert g-h rvert^2  mathrm{d} alpha right}^{1/2}?$$ This does not looks like the proper usage of the "Schwarz inequality" or am I missing something? Can someone please propose a correct version of the solution?
The solution to the same question here seems to suffer from the same issue, if this is an issue at all, that is.

Zeta-Squared · Accepted Answer

Regarding your first question I agree, it seems it should be $|g-h|_{2}^{2}$ on the LHS.
For your second question, the Cauchy-Schwarz inequality is a special case of H$ddot{text{o}}$lder's inequality in the sense that it is the case that is applicable to inner products. So in the second display you have which explains this the intermediate steps are,
begin{align}
bigg|int_{a}^{b}fgdalphabigg|leqint_{a}^{b}|f||g|dalpha=|fg|_{1}leq|f|_{2}|g|_{2}=bigg(int_{a}^{b}|f|^{2}dalphabigg)^{1/2}bigg(int_{a}^{b}|g|^{2}dalphabigg)^{1/2}
end{align}
The other thing to note here is it seems you are misreading the solution. You are asking how,
begin{align}
int_{a}^{b}|f-g||g-h|dalphaleq|f|_{2}|g|_{2},
end{align}
which is not what is being presented in the proof. In Rudin's proof he is using,
begin{align}
int_{a}^{b}|f-g||g-h|dalphaleq|f-g|_{2}|g-h|_{2}.
end{align}
EDIT: To clarify the last inequality, since $f,g,h$ are all Riemann integrable with respect to $dalpha$ then $f-g$ and $g-h$ are also Riemann integrable. So consider,
begin{align}
int_{a}^{b}|f-g||g-h|dalpha=|(f-g)(g-h)|_{1}leq|f-g|_{2}|g-h|_{2}=bigg(int_{a}^{b}|f-g|^{2}dalphabigg)^{1/2}bigg(int_{a}^{b}|g-h|^{2}dalphabigg)^{1/2}
end{align}
EDIT2: As Paramanand pointed out, since $f-g$ and $g-h$ are Riemann integrable replace $f$ by $|f-g|$ and $g$ by $|g-h|$ in the given Cauchy-Schwarz inequality to get,
begin{align}
int_{a}^{b}|f-g||g-h|dalpha=bigg|int_{a}^{b}|f-g||g-h|dalphabigg|leqbigg(int_{a}^{b}|f-g|^{2}dalphabigg)^{1/2}bigg(int_{a}^{b}|g-h|^{2}dalphabigg)^{1/2}
end{align}

Possible mistake in the solution of Baby Rudin Ch. 6 Ex. 11

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