I'm working through Spivak's Calculus over the summer, and I'm currently on problem 19 of Chapter 1, which involves proving the Schwarz inequality. The first two parts of the proof are fairly straightforward, but I don't understand the last part. Spivak gives the inequality:


and suggests using the quadratic formula (I'm not sure I understand this part, since the equation is greater than zero, how can there be any solutions?) to arrive at the Schwarz inequality. In the answer key, the following is given.


I'm not really sure how this follows since I can't get it just by trying to find the roots of the equations vis a vis the quadratic formula.

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    $\begingroup$ For the first inequality look at the right hand side as a quadratic expression in terms of $\lambda$ and take the discriminant: $b^2-4ac$. Can you show that it is always negative? Then it means that it never becomes zero. Now, what is the coefficient of the $\lambda ^2$? And what does that tell you? $\endgroup$ – Keivan May 20 '12 at 20:23

Your observation that there is no solution is precisely the key to the solution.

You have $$0=\lambda^2(y_1^2+y_2^2)-2\lambda(x_1y_1+x_2y_2)+(x_1^2+x_2^2)$$

has a real solution if and only if the discriminant (the part under the root in the quadratic formula) is $\geq 0$.

But the discriminant is just


Putting the pieces together this yields: Since the equation has no solution, the discriminant is negative. Rearranging gives Cauchy-Schwarz.

Edit: I assume you know how to get the first inequality, but just for completeness note that it is equivalent to $\left<\lambda y-x,\lambda y-x\right>>0$. And to be totally precise you have to deal with the case that $x$ and $y$ are linearly dependent separately.

  • $\begingroup$ Oh. Awesome! I had trouble stepping back to see that. $\endgroup$ – user17137 May 20 '12 at 20:34

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