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I am following the course of Differential Equations offered in the OCW (https://ocw.mit.edu/courses/mathematics/18-034-honors-differential-equations-spring-2009/index.htm).

In the Book of Ordinary Differential Equations, by Garrett Birkhoff and Gian-Carlo Rota, while proving the uniqueness Theorem for Differential equations of second order (4th edition - p.42), they do the following:

$$E'(x)=-2p(x)u'^2+2[1-q(x)]uu' $$

Since $(u\pm u')^2\ge0$, it follows that $\lvert2uu'\rvert \le u^2 + u'^2)$. Hence $$2[1-q(x)]uu'\le (1+\lvert q(x)\rvert)(u^2+u'^2 )$$

and $$E'(x)\le[1+\lvert q(x) \rvert]u^2 +[1+\lvert q(x) \rvert + \lvert 2p(x) \rvert]u'^2$$

I do not understand why do they use the absolute value... wouldn't it be right if they used only:

$$2[1-q(x)]uu'\le (1-q(x))(u^2+u'^2 )$$

Checking in the course notes, it is said that the Cauchy Schwarz inequality is used. However, I do not understand how does the Cauchy Schwarz inequality justifies the use of the absolute value.

I would really appreciate if you could help me

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    The inequality you are claiming would be true if you knew that $1-q(x)$ were positive. Since (presumably) this isn't given, the best you can say is to put absolute values on the right-hand side, i.e. have something like $|1-q(x)|(u^2+u'^2)$. The authors replaced $|1-q(x)|$ by $1+|q(x)|$, since it is true that $|1-q(x)|\le 1+|q(x)|$ by the triangle inequality. – 2017-01-03
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    Thanks for your answer, it really helped me a lot. For those who want more information about the triangle inequality, you can check "Mathematical expression of the constraint on the sides of a triangle" in https://en.wikipedia.org/wiki/Triangle_inequality. – 2017-01-12

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