TITLE: Spivak Chapter 1, Problem 19(a)
QUESTION [5 upvotes]: Per Spivak

The great granddaddy of all inequalities is the Schwarz inequality $${x}_{1}{y}_{1}+{x}_{2}{y}_{2} \le \sqrt{{x}_{1}^{2} + {x}{_2}^{2}}\sqrt{{y}_{1}^{2} + {y}{_2}^{2}}$$


Prove that if ${x}_{1} = \lambda {y}_{1}$ and ${x}_{2} = \lambda {y}_{2}$ for some number $\lambda \ge 0$, then equality holds in the Schwarz inequality.

There is an answer here, but it seems to follow the pattern elsewhere of simply substituting on both sides of the given inequality, and changing the $\le$ to an $=$, unless I am misunderstanding it. Amusingly, having invested in Spivak's answer book, his answer is "The proofs for .... are straightforward" :-)
Here is my attempt.
Given ${x}_{1} = \lambda y_{1}$ and ${x}_{2} = \lambda y_{2}$ then
$$\sqrt{{x}_{1}^{2} + {x}_{2}^{2}}\sqrt{{y}_{1}^{2} + {y}_{2}^{2}} = \sqrt{(\lambda {y}_{1})^{2} + (\lambda {y}_{2})^{2}}\sqrt{{y}_{1}^{2} + {y}_{2}^{2}}$$
$$= \sqrt{\lambda^{2} {y}_{1}^{2} + \lambda^{2} {y}_{2}^{2}}\sqrt{{y}_{1}^{2} + {y}_{2}^{2}}$$
$$=\sqrt{\lambda^{2}( {y}_{1}^{2} +  {y}_{2}^{2})}\sqrt{{y}_{1}^{2} + {y}_{2}^{2}}$$
$$=\sqrt{\lambda^{2}}\sqrt{( {y}_{1}^{2} +  {y}_{2}^{2})}\sqrt{{y}_{1}^{2} + {y}_{2}^{2}}$$  Note: $\sqrt {\lambda^{2}} = \lambda \iff \lambda \ge 0 $ as asserted in the problem.
$$=\lambda ({y}_{1}^{2} +  {y}_{2}^{2})$$
$$=\lambda {y}_{1}^{2} +  \lambda{y}_{2}^{2}$$
$$=\lambda {y}_{1}{y}_{1} +  \lambda{y}_{2}{y}_{2}$$
$$=(\lambda {y}_{1}){y}_{1} +  (\lambda{y}_{2}){y}_{2}$$
$$=x_{1}{y}_{1} +  x_{2}{y}_{2}$$

REPLY [2 votes]: I think the exact problem should be like this,
$$|x_{1}y_{1}+x_{2}y_{2}|\le \sqrt{x_{1}^{2}+x_{2}^{2}}\sqrt{y_{1}^{2}+y_{2}^{2}}$$
If $x_{1}=\lambda y_1$ and $x_2=\lambda y_2$
Now it's obvious that,
$$|x_{1}y_{1}+x_{2}y_{2}|\le |x_{1}y_{1}|+|x_{2}y_{2}|$$
... After following what op done in the description of the problem, we are left with,
$$|x_{1}y_{1}+x_{2}y_{2}|\le |x_{1}y_{1}|+|x_{2}y_{2}|\le \sqrt{x_{1}^{2}+x_{2}^{2}}\sqrt{y_{1}^{2}+y_{2}^{2}}$$
Which implies that,
$$|x_{1}y_{1}+x_{2}y_{2}|\le \sqrt{x_{1}^{2}+x_{2}^{2}}\sqrt{y_{1}^{2}+y_{2}^{2}}$$