# Cauchy-Schwarz Inequality for Complex Numbers

Simple question: do we really need the conjugate in the inequality?

$$|\sum_{j=1}^n a_j \overline{b_j}|^2 \leq \sum_{j=1}^n |a_j|^2 \sum_{j=1}^n |b_j|^2$$

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The answer is No. –  Tom Nov 17 '12 at 12:31

The Cauchy-Schwarz inequality says that $|\langle a,b \rangle| \leq \| a \| \|b\|$. In $\mathbb C^n$, the inner product is $$\langle a, b \rangle = \sum_{j=1}^n a_j \bar{b_j}.$$

That's why we the Cauchy-Schwarz inequality in $\mathbb C^n$ has conjugates in it.

While it is true that you could omit the conjugates in your inequality and still have a true statement, that would only take us further away from the nice statement that $|\langle a,b \rangle| \leq \| a \| \|b\|$.

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Ok thanks for the answer. Haven't read about the inner product for complex numbers yet –  Dan Nov 17 '12 at 12:44

Simple answer, yes. Look back to the definition of the inner product (for complex numbers) and note the complex conjugate symmetry (as opposed to pure symmetry)

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If you replace $b_j$ by $\overline{b_j}$ on both sides, you have no more $\overline{.}$ on the left and on the right, you don't have any either since you have $|b_j| = |\overline{b_j}|$... –  xavierm02 Nov 17 '12 at 12:33