What is the name of this inequality?

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My prof. calls it the triangle inequality. However the wikipedia page with the same this name shows a special case of it, which is ##|x+y|\leq|x|+|y|##, and my prof. calls it the triangle inequality 2. I wonder what the formal name of the inequality in the picture above is. Thanks in adv.
 
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|(x-z)+(z-y)| \le |x-z|+|z-y|
 
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Leo Liu said:
Homework Statement:: .
Relevant Equations:: .

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My prof. calls it the triangle inequality. However the wikipedia page with the same this name shows a special case of it, which is ##|x+y|\leq|x|+|y|##, and my prof. calls it the triangle inequality 2. I wonder what the formal name of the inequality in the picture above is. Thanks in adv.
Triangle inequality is correct. The reason why is in post #2 by @anuttarasammyak .
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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