A sub-additivity inequality
In trying to understand a result of D. Rider (Trans. AMS, 1973) I've got stuck on a lemma that he uses. At one point he makes a step without comment or explanation, but I can't see why it works.
Here is a paraphrase of what I think he is claiming. Let $d_1,\dots, d_n$ be complex numbers of modulus $1$; let $b_1,\dots, b_n$ be complex numbers of modulus $\leq 1$. (In the intended application the numbers $b_1,\dots, b_n$ are actually the diagonal entries of a unitary $n\times n$ matrix, but I strongly suspect that is not used in the argument.) Then Rider seems to assert, without further comment, that $$ \left\vert\sum_{j=1}^n (d_jb_j-1)\right\vert^{1/2} \leq \left\vert\sum_{j=1}^n (d_j-1)\right\vert^{1/2} + \left\vert\sum_{j=1}^n (b_j-1)\right\vert^{1/2} $$
Probably I am just being dense, and have failed to spot a routine estimate that does the job. If anyone could get me started on the right track that would be most helpful, as this is starting to get very irritating, and is not even the main part of Rider's argument.
Update Firstly, thanks to everyone who answered, but in particular to Will Jagy for various off-MSE exchanges, and to George Lowther for his elegant argument (the key part that I had failed to think of, was the use of $\rm Re$ and its linearity rather than $\vert\cdot\vert$ and its sub-additivity, allowing one to boost up the observation made by Gerry Myerson).
In case it's of interest, here are some more details on how the question I asked corresponds to Lemma 5.1 in Rider's paper. The paper is dealing with central trigonometric polynomials on compact groups, which means: linear combinations of traces of irreducible representations. Paraphrased lightly, and with a fairly trivial reduction step thrown in, the aforementioned lemma says the following:
Let $\phi: G \to U(n)$ be an (irreducible) unitary representation of a (compact) group $G$. If $$\vert n^{-1}\operatorname{\rm Tr}\phi(g_i)-z_i\vert \leq \delta_i \qquad(i=1,\dots, p)$$ where $\vert z_i\vert=1$ for all $i$, then $$ \left\vert n^{-1}\operatorname{Tr}\phi(g_1\dotsb g_n) - \prod_{i=1}^p z_i \right\vert \leq \left(\sum_{i=1}^p \delta_i^{1/2}\right)^2 $$
The proof given in the paper goes as follows: reduce to the case $p=2$ (if this can be done then induction will do the rest); then observe that WLOG $\phi(g_1)$ is a diagonal unitary matrix, w.r.t. an appropriate basis. If we let $d_1,\dots, d_n$ be the diagonal entries of $\phi(g_1)$ and $b_1,\dots, b_n$ be those of $\phi(g_2)$, then the trace of $\phi(g_1g_2)$ is just $\sum_{i=1}^n d_ib_i$, and so we have to prove
Claim. If $|z_1|=|z_2|=1$, $\left\vert\sum_{i=1}^n (d_i - z_1)\right\vert\leq n\delta_1$, and $\left\vert\sum_{i=1}^n (b_i - z_2)\right\vert\leq n\delta_2$, then $$\left\vert \sum_{i=1}^n (d_ib_i - z_1z_2) \right\vert \leq n(\delta_1^{1/2}+\delta_2^{1/2})^2. $$
This is where Rider is content to say "done"; and it is hopefully clear that this is equivalent to my original question.
Yes, the result holds. First, for any complex number $z$ with $\vert z\vert\le1$, we have $$ 2\Re(1-z)=1-\vert z\vert^2+\vert z-1\vert^2\ge\vert z-1\vert^2. $$ It follows that if $Z$ is a complex random variable with $\vert Z\vert\le1$ then, $$ \mathbb{E}\left[\left\vert Z-1\right\vert^2\right]\le2\Re\left(\mathbb{E}\left[1-Z\right]\right)\le2\left\vert\mathbb{E}\left[Z-1\right]\right\vert. $$ Therefore, if $X,Y$ are complex random variables with $\vert X\vert\le1$ and $\vert Y\vert\le1$ then, using Cauchy–Schwarz, $$ \begin{align} \left\vert\mathbb{E}\left[(X-1)(Y-1)\right]\right\vert^2&\le\mathbb{E}\left[\left\vert X-1\right\vert^2\right]\mathbb{E}\left[\left\vert Y-1\right\vert^2\right]\cr &\le4\left\vert\mathbb{E}\left[X-1\right]\mathbb{E}\left[Y-1\right]\right\vert \end{align} $$ So, using the identity $XY-1=(X-1)+(Y-1)+(X-1)(Y-1)$ (as in Gerry Myerson's answer), $$ \begin{align} \left\vert\mathbb{E}\left[XY-1\right]\right\vert&\le\left\vert\mathbb{E}\left[X-1\right]\right\vert+\left\vert\mathbb{E}\left[Y-1\right]\right\vert+\left\vert\mathbb{E}\left[(X-1)(Y-1)\right]\right\vert\cr &\le\left\vert\mathbb{E}\left[X-1\right]\right\vert+\left\vert\mathbb{E}\left[Y-1\right]\right\vert+2\left\vert\mathbb{E}\left[X-1\right]\mathbb{E}\left[Y-1\right]\right\vert^{1/2}\cr &=\left(\left\vert\mathbb{E}[X-1]\right\vert^{1/2}+\left\vert\mathbb{E}[Y-1]\right\vert^{1/2}\right)^2. \end{align} $$ Take the square root of each side to obtain $$ \left\vert\mathbb{E}\left[XY-1\right]\right\vert^{1/2}\le\left\vert\mathbb{E}[X-1]\right\vert^{1/2}+\left\vert\mathbb{E}[Y-1]\right\vert^{1/2}, $$ which is equivalent to the inequality in the question. Note that it is only necessary to have $\vert d_i\vert\le1$ and $\vert b_i\vert\le1$ (we do not need $\vert d_i\vert=1$).
Well, I can do the case $n=1$.
$$|db-1|=|(d-1)+(b-1)+(d-1)(b-1)|\le|d-1|+|b-1|+|(d-1)(b-1)|$$ and $$|(d-1)(b-1)|\le2|(d-1)(b-1)|^{1/2}$$ taken together and taking the square root on both sides gives $$|db-1|^{1/2}\le|d-1|^{1/2}+|b-1|^{1/2}$$
Bill Johnson asked about truth for other exponents, and I finally realize that the exponent $1/2$ really is correct and not arbitrary. $$ \left| \sum_{j=1}^n (d_jb_j-1)\right|^\alpha \leq \left| \sum_{j=1}^n (d_j-1)\right|^\alpha + \left| \sum_{j=1}^n (b_j-1)\right|^\alpha $$ is false for some data with $n=2$ as soon as $\alpha > 1/2.$ For some $0 \leq \theta \leq \pi/2,$ take $$ d_1 = b_1 = e^{i \theta}, \; \; d_2 = b_2 = \bar{d_1} = \bar{b_1} = e^{-i \theta}. $$ Then the inequality with exponent $\alpha$ is false for $$ 0 < \theta < \arccos \left( \frac{2^{1/\alpha} - 2}{2} \right). $$ For $\alpha = 1,$ which is my previous "answer," I used $\theta = \pi/3$ as optimal in one sense, but I could have used any $0 \leq \theta \leq \pi/2.$
$\theta = \pi/3$ gives falsehood as long as $\alpha > \frac{\log 2}{\log 3} \approx 0.6309...$
As $\alpha$ decreases to $1/2,$ the window for $\theta$ becomes narrower, until finally we get to $\alpha = 1/2$ and $0 < \theta < \arccos 1 = 0,$ no such thing.