RTP: $${\displaystyle |\langle \mathbf {u} ,\mathbf {v} \rangle |\leq \|\mathbf {u} \|\|\mathbf {v} \|.}$$

I know there are many ways to prove the Cauchy-Schwarz inequality, however most proofs that I've been looking at start with $$({a\vec{x}-b\vec{y}})^2$$, where a,b ∈ℝ and x,y ∈ 𝕍. The rest of the proof is quite easy to follow once the values for a and b are chosen, my question is why did we start off with $$({a\vec{x}-b\vec{y}})^2$$, where did it come from, and what constitutes our choice for the coefficients a and b.

  • $\begingroup$ math.stackexchange.com/questions/2698335/… $\endgroup$
    – user284331
    Mar 20, 2018 at 18:32
  • $\begingroup$ $(a \vec x - b \vec y)^2$ is a natural quantity to consider because it unites the norms of $x$ and $y$ with $\langle x, y\rangle$. $\endgroup$
    – user296602
    Mar 20, 2018 at 18:33
  • $\begingroup$ I dont know if it applies to this proof but, in general, the proof of the Cauchy-Schwarz inequality use orthogonal projections. $\endgroup$
    – Masacroso
    Mar 20, 2018 at 18:35
  • $\begingroup$ Schwarz's genius. $\endgroup$
    – user123641
    Mar 20, 2018 at 18:36
  • $\begingroup$ @Masacroso can you please elaborate on this, I was trying to find a link geometrically but it only makes sense for second and third real dimension, however I am trying to prove the inequality for R^n, so how would projections relate to the proof ? $\endgroup$
    – Cbb7
    Mar 20, 2018 at 18:41

3 Answers 3


We start with $\|ax-by\|^2$ since it is a non-negative number, with the hope to derive the Cauchy-Schwarz inequality from a trivial inequality. For instance, by just picking $a$ and $b$ as $1$ we have $$0 \leq \|x-y\|^2 = \|x\|^2+\|y\|^2 -2\langle x,y\rangle \tag{1}$$ from which it follows that $$ \langle x,y\rangle \leq \frac{\|x\|^2+\|y\|^2}{2}\qquad\text{(trivial inequality)}.\tag{2}$$ Now we may exploit a symmetry: if we pick some $\lambda>0$ and replace $x$ by $\lambda x$ and $y$ by $\frac{1}{\lambda}y$ the LHS of $(2)$ is unchanged. In particular $$ \forall \lambda>0,\qquad\langle x,y\rangle \leq \frac{\lambda^2\|x\|^2+\frac{1}{\lambda^2}\|y\|^2}{2}\qquad\text{(less trivial inequality)}.\tag{3}$$ Given some $\|x\|,\|y\|>0$, we may pick $\lambda>0$ such that the RHS of $(3)$ is minimal.
By the AM-GM inequality the RHS of $(3)$ is $\geq\|x\|\|y\|$, but equality is attained if $\lambda=\sqrt{\frac{\|y\|}{\|x\|}}$. $$ \langle x,y\rangle \leq \|x\|\|y\| \tag{4} $$ Is the Cauchy-Schwarz inequality. By performing a backtracking in the outlined proof we have that equality holds iff $x,y$ are linearly dependent. We also have that $(4)$ holds for any positive definite inner product $\langle \cdot,\cdot\rangle$, where $\|x\|$ is defined through $\|x\|^2=\langle x,x\rangle$.

  • $\begingroup$ Thank you, it's a simple, yet an amazing approach. $\endgroup$
    – Cbb7
    Mar 20, 2018 at 19:01
  • $\begingroup$ @Cbb7: I will try to make it clearer: by the AM-GM inequality, the RHS of $(3)$ is $\geq\|x\|\|y\|$ for any $\lambda>0$. On the other hand, by picking $\lambda$ as $\sqrt{\|y\|/\|x\|}$ the RHS of $(3)$ is exactly equal to $\|x\|\|y\|$, hence $\|x\|\|y\|$ is the minimum of the RHS of $(3)$ as $\lambda$ ranges over $\mathbb{R}^+$. This proves $(3)\to(4)$. $\endgroup$ Mar 20, 2018 at 19:02
  • $\begingroup$ yeppp, I missed the fact that λ^2=‖y‖‖x‖, again thanks a lot for your help $\endgroup$
    – Cbb7
    Mar 20, 2018 at 19:03

You'd be surprised how many proofs of inequalities, be they with real numbers or vectors (often it comes down to the same thing), boil down to the simple fact that certain kinds of square (such as $x^2,\,x\in\mathbb{R}$ or $\langle x,\,x\rangle$) cannot be negative. In the case of inner products, that's just about the only basic constraint on them; in fact it's an axiom. So if you want to find the tightest constraint you can, you have to restate the problem as, "find out when this non-negative quadratic function is zero".


In order to prove $${\displaystyle |\langle \mathbf {u} ,\mathbf {v} \rangle |\leq \|\mathbf {u} \|\|\mathbf {v} \|.}$$

Note that

If $$ ax^2 +bx +c \ge 0 $$ for all values of $x$ then we have $$ b^2 \le 4ac$$ Now we try to turn

$${\displaystyle |\langle \mathbf {u} ,\mathbf {v} \rangle |^2\leq \|\mathbf {u} \|^2\|\mathbf {v} \|^2.}$$

into $$ b^2 \le 4ac $$

by choosing the correct a and b.

Thus we have to complete the square and $$({a**x**-b**y**})^2 $$ shows up.


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