# Proof of Cauchy-Schwartz inequality with dot product and euclidean norm

I have some problems on understanding the proof of Cauchy-Schwartz inequality from my textbook:

Given $$\textbf{x,y} \in \mathbb{R} \Rightarrow \vert \textbf{x}^T \textbf{y} \vert \le \Vert \textbf{x} \Vert_2 \cdot \Vert \textbf{y} \Vert_2$$

the proof:

given $$\lambda \in \mathbb{R}$$ we can observe that : \begin{align} 0 \le \Vert \textbf{x} + \lambda \textbf{y} \Vert_2^2 & = \sum_{i=1}^n (x_i + \lambda y_i)^2 \\ & = \sum_{i=1}^n (x_i^2 + 2 \lambda x_i y_i + \lambda^2 y_i^2) \\ & = \sum_{i=1}^n x_i^2 + 2\lambda \sum_{i=1}^n x_iy_i + \lambda^2 \sum_{i=1}^n y_i^2 \\ & = \Vert \textbf{x} \Vert^2_2 + 2 \lambda \textbf{x}^T \textbf{y} + \vert \lambda \vert^2 \Vert \textbf{y} \Vert^2_2 \end{align} if $$\textbf{x}^T\textbf{y} = 0$$ then the thesis is surely true.
Instead, if $$\textbf{x}^T\textbf{y} \ne 0$$, then we can consider: $$\lambda = - \frac{\Vert \textbf{x} \Vert^2_2}{\textbf{x}^T\textbf{y}}$$ therefore we have: \begin{align} 0 \le \Vert \textbf{x} \Vert^2_2 - 2\Vert \textbf{x} \Vert^2_2 + \frac{\Vert \textbf{x} \Vert^4_2}{\vert \textbf{x}^T \textbf{y} \vert^2} \Vert \textbf{y} \Vert^2_2 & = -\Vert \textbf{x} \Vert^2_2 \frac{\Vert \textbf{x} \Vert^4_2}{\vert \textbf{x}^T \textbf{y} \vert^2} \Vert \textbf{y} \Vert^2_2 \\ & = \Vert \textbf{x} \Vert^2_2 \left ( -1 + \frac{\Vert \textbf{x}\Vert^2_2 \Vert \textbf{y}\Vert^2_2 }{ \vert \textbf{x}^T \textbf{y} \vert^2 }\right ) \end{align} we can deduce that : $$\Vert \textbf{x} \Vert^2_2 \Vert \textbf{y} \Vert^2_2 - \vert \textbf{x}^T \textbf{y} \vert^2 \ge 0$$ and then the thesis follows easily.

There are some questions I want to post here:

1) why immediately is it observed that $$0 \le \Vert \textbf{x} + \lambda \textbf{y} \Vert_2^2$$, from where we can deduce that observation?

2) when it says << if $$\textbf{x}^T\textbf{y} = 0$$ then the thesis is surely true. >> for "thesis", does it mean the expression of above proposition?:

$$\vert \textbf{x}^T \textbf{y} \vert \le \Vert \textbf{x} \Vert_2 \Vert \textbf{y} \Vert_2$$

but if I substitute in the last passage:

\begin{align}& = \Vert \textbf{x} \Vert^2_2 + 2 \lambda \textbf{x}^T \textbf{y} + \vert \lambda \vert^2 \Vert \textbf{y} \Vert^2_2 \\ & = \Vert \textbf{x} \Vert^2_2 + 2 \lambda 0 + \vert \lambda \vert^2 \Vert \textbf{y} \Vert^2_2 \\ & = \Vert \textbf{x} \Vert^2_2 + \vert \lambda \vert^2 \Vert \textbf{y} \Vert^2_2 \end{align}

I do not obtain the $$\Vert \textbf{x} \Vert_2 \Vert \textbf{x} \Vert_2$$ of the thesis of the proposition.

3) why considering the $$\lambda$$ in that way?

Instead, if $$\textbf{x}^T\textbf{y} \ne 0$$, then we can consider $$\lambda = - \frac{\Vert \textbf{x} \Vert^2_2}{\textbf{x}^T\textbf{y}}$$

Please, can you help me to understand better? Many thanks!

• (1) From the definition of inner product and the norm induced from it $\|x\|^2=(x,x)\geq0$. (2) Thesis is what you want to prove, yes. (3) If you look at that polynomial in $\lambda$, which is quadratic, that specific value is the vertex of the parabola, at which you attain the smallest value. Since you know that all values are $\geq0$, it follows that the smallest value is also $\geq0$. You choose the smallest value to get a result as tight as possible, and also symmetric in $x$ and $y$. – user647486 Mar 16 at 19:28
• That last part of the proof is more natural if instead of telling which value of $\lambda$ to take, you complete the square, or try to minimize the quadratic polynomial by taking derivatives with respect to $\lambda$. But some authors prefer to write math like a magic show instead of experiment and deduction, which is how mathematics is really created. – user647486 Mar 16 at 19:36
• @user647486 can explain better the (1)? note that I have corrected the typo in the statement. – JB-Franco Mar 16 at 19:49
• There is no change in the explanation of (1). By definition of inner product $(x,x)\geq0$. Ah! In your notation and particular case of finite dimension $x^Tx\geq0$ just because that is a sum of squares of real numbers (the coordinates of $x$). Then $\|x\|^2=x^Tx$. – user647486 Mar 16 at 19:59

(1) This is from the definition of a norm; namely, they are nondegenerate. (2) The RHS is non-negative (again by nondegeneracy) and so if the LHS is 0, then the inequality must be satisfied. (3) This choice of $$\lambda$$ makes the arithmetic come out like you want. He proves a general inequality and then picks a specific $$\lambda$$ so that the more general inequality will reduce to the desired inequality.
• in (3) what is the way to follow to deduce that $\lambda$? I tried to obtain $\lambda$ from the previous inequality passage but without success. – JB-Franco Mar 16 at 20:12
• I'm not sure if there's an intuitive way to explain how to come up with that $\lambda$. I imagine if you did the computation he wrote up, but just left $\lambda$ in the equation, then you'd come to a point where you should see that that choice of $\lambda$ will give the necessary cancellation to properly simplify the inequality. – Gary Moon Mar 16 at 21:53
1. It follows from the fact the a norm is always greater than or equal to $$0$$.
2. The statement of the Cauchy-Schwarz equality is not what you wrote. It is $$\lvert x^Ty\rvert\leqslant\lVert x\rVert_2\lVert y\rVert_2$$. This obviously holds if $$x^Ty=0$$.