I don't grasp some steps in this quick proof of Cauchy-Schwarz inequality.

Let $\boldsymbol{u}$ and $\boldsymbol{v}$ be two vectors in $\mathbb{R}^n$. The Cauchy-Schwarz inequality states that $$ \boldsymbol{u} \cdot \boldsymbol{v} \leq |\boldsymbol{u}||\boldsymbol{v}| $$

We know that \begin{align} |\boldsymbol{u}+\boldsymbol{v}|^2&=(\boldsymbol{u}+\boldsymbol{v})\cdot (\boldsymbol{u}+\boldsymbol{v})\\ &= \boldsymbol{u}\cdot \boldsymbol{u}+\boldsymbol{v} \cdot \boldsymbol{v} +2\boldsymbol{u}\cdot \boldsymbol{v}\\ &=|\boldsymbol{u}|^2+|\boldsymbol{v}|^2+2\boldsymbol{u}\cdot \boldsymbol{v} \tag{1} \end{align} Using Cauchy-Schwarz, we have $$ |\boldsymbol{u}|^2 +|\boldsymbol{v}|^2+2\boldsymbol{u}\cdot \boldsymbol{v} \leq |\boldsymbol{u}|^2+|\boldsymbol{v}|^2+2|\boldsymbol{u}||\boldsymbol{v}|=(|\boldsymbol{u}|+|\boldsymbol{v}|)^2 \tag{2} $$ So the Cauchy-Schwarz inequality gives $$ |\boldsymbol{u}+\boldsymbol{v}|^2 \leq (|\boldsymbol{u}|+|\boldsymbol{v}|)^2 $$ or $$ |\boldsymbol{u}+\boldsymbol{v}| \leq (|\boldsymbol{u}|+|\boldsymbol{v}|) $$

I don't understand the inequality in equation (2). How can we conclude the inequality? I.e why is $$ |\boldsymbol{u}|^2 +|\boldsymbol{v}|^2+2\boldsymbol{u}\cdot \boldsymbol{v} \leq |\boldsymbol{u}|^2+|\boldsymbol{v}|^2+2|\boldsymbol{u}||\boldsymbol{u}| \quad\text{?} $$

And how is $\boldsymbol{u} \cdot \boldsymbol{v} \leq |\boldsymbol{u}||\boldsymbol{v}|$ equivalent to $ |\boldsymbol{u}+\boldsymbol{v}| \leq (|\boldsymbol{u}|+|\boldsymbol{v}|)$?

  • $\begingroup$ This is not a proof of the inequality, it is a proof that uses it. $\endgroup$ – user416426 Feb 16 '17 at 19:49
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    $\begingroup$ Remember that $u.v$ is the product of the norm and the cosinus of the angle (which is bounded by $1$). $\endgroup$ – anonymus Feb 16 '17 at 19:51
  • $\begingroup$ see here i hope this will help you en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality $\endgroup$ – Dr. Sonnhard Graubner Feb 16 '17 at 19:53

This is a proof of the triangle inequality $|u+v| \leqslant |u| + |v|$ that depends on the truth of the Cauchy-Schwarz inequality $u \cdot v \leqslant |u| |v|$. This proof is not proving the Cauchy-Schwarz inequality, it is using the validity of the Cauchy-Schwarz inequality in order to prove something else.


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