You are correct... to an extent. When you open up the program, it says the image is 5832 pixels wide and 300 dpi. Thus it's 5832/300 = 19.44 inches wide. If you then add your 1.75" @ 300dpi, you are correct that you've effectively left the pic alone and just added extra border (in this case, it would be 525 pixels (300 x 1.75) all the way around). BUT, that
only works if you plan to print it at 19.44" wide, what I would call the native/original size.
But let's assume that Slay wants to print 32" wide plus a border. If you leave it at 300dpi, and stretch it to 32", the image grows to 32x300 = 9600 pixels wide. But nearly half of those pixels are interpolations, padded inserts that usually just average the color values of the pixels on either side, making it higher res but blurred. Then when you save it, the JPEG algorithm has to account for those new pixels as new information, but they really don't add any new information entropy-wise. The file size increases, you have a chance of getting some artifacting depending on the compression level. And every time you save it, it will get worse because jpeg is a "lossy" compression (i.e. it actually discards information every time you save it).
You're better off keeping it 5832 pixels wide and changing the resolution to 182 dpi (5832/32 ~= 182). Then when you add 1.75", you're only adding 319 pixels (182 x 1.75) all the way around.
So if you're keeping the number of pixels in the main image fixed, the number of pixels needed for the 1.75" border will vary based on the dpi, which is determined by the size of the physical printing. It's a cleaner way to do it. I would only use interpolation if you were making it so big that the dpi dropped below 100-150. And then I would use fractal interpolation that is far smarter at adding the pixels than just mere averaging. It keeps the edges and other little details much sharper.
Bottom line, for minor changes, you probably can't see the difference and I'm just being super particular (I'm a picky SOB). But if you do much image work, the changes all add up and eventually degrade the quality of your image. I could go full-nerd and dive into 2-D discrete fourier transforms and information theory about the inner workings of the jpeg algorithm, but I'm tired and you're going to have to take my word for it
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A much simpler way to look at your original question (without all my rambling). If you print this image @ 10" wide, a 1.75" border would be much larger proportionally than a 1.75" border on an image 30" wide. Make sense?