<s>[INST] Verifying that $\left[\int_a^b(x-p)\,dx\right]^2\le(b-a)\int_a^b(x-p)^2\,dx$ without Cauchy-Schwarz 
For real numbers $a,b,p$, verify that $$\left[\int_a^b(x-p)\,dx\right]^2\le(b-a)\int_a^b(x-p)^2\,dx\tag{1}$$ without using the Cauchy-Schwarz inequality.

Note that this is obviously an immediate consequence of C-S, and I assumed that it had to be shown by integrating directly.
Working backwards, we get $$\left[\frac{b^2-a^2}2-p(b-a)\right]^2\le(b-a)\frac{(b-p)^3-(a-p)^3}3$$ and since $m^3-n^3=(m-n)(m^2+mn+n^2)$, the RHS can be rewritten as $$\frac{(b-a)^2}4(b+a-2p)^2\le\frac{(b-a)^2}3(b^2+ab+b^2+3p^2-3p(b+a))$$ or $$3(b+a-2p)^2\le4((b+a)^2+3p^2-3p(b+a))-4ab$$ which gives $$12p^2-12p(b+a)\le(b-a)^2+12p^2-12p(b+a)$$ or that $$(b-a)^2\ge0$$ which is true.
Any alternative methods?
 [/INST] Do the substitution $t=x-p$, so the inequality becomes
$$
\left(\int_{a-p}^{b-p} t\,dt\right)^{\!2}\le (b-a)\int_{a-p}^{b-p} t^2\,dt
$$
Setting $a-p=A$ and $b-p=B$, this is the same as proving that
$$
\left(\int_{A}^{B} t\,dt\right)^{\!2}\le (B-A)\int_{A}^{B} t^2\,dt
$$
that is,
$$
\frac{(B^2-A^2)^2}{4}\le (B-A)\frac{B^3-A^3}{3}
$$
that becomes
$$
3(B+A)^2\le 4(B^2+AB+A^2)
$$
This becomes
$$
A^2-2AB+B^2\ge0
$$
which is true.
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