Cauchy–Schwarz inequality

Sedrakyan's inequality, also called Bergström's inequality, Engel's form, the T2 lemma, or Titu's lemma, states that for real numbers ${\displaystyle u_{1},u_{2},\dots ,u_{n}}$ and positive real numbers ${\displaystyle v_{1},v_{2},\dots ,v_{n}}$:

$${\displaystyle {\frac {\left(\sum _{i=1}^{n}u_{i}\right)^{2}}{\sum _{i=1}^{n}v_{i}}}\leq \sum _{i=1}^{n}{\frac {u_{i}^{2}}{v_{i}}}\quad {\text{ or equivalently, }}\quad {\frac {\left(u_{1}+u_{2}+\cdots +u_{n}\right)^{2}}{v_{1}+v_{2}+\cdots +v_{n}}}\leq {\frac {u_{1}^{2}}{v_{1}}}+{\frac {u_{2}^{2}}{v_{2}}}+\cdots +{\frac {u_{n}^{2}}{v_{n}}}.}$$

It is a direct consequence of the Cauchy–Schwarz inequality, obtained by using the dot product on ${\displaystyle \mathbb {R} ^{n}}$ upon substituting ${\displaystyle u_{i}'=u_{i}/{\sqrt {v_{i}}}}$ and ${\displaystyle v_{i}'={\sqrt {v_{i}}}}$. This form is especially helpful when the inequality involves fractions where the numerator is a perfect square.

Cauchy-Schwarz inequality in a unit circle of the Euclidean plane

The real vector space ${\displaystyle \mathbb {R} ^{2}}$ denotes the 2-dimensional plane. It is also the 2-dimensional Euclidean space where the inner product is the dot product. If ${\displaystyle \mathbf {u} =\left(u_{1},u_{2}\right)}$ and ${\displaystyle \mathbf {v} =\left(v_{1},v_{2}\right)}$ then the Cauchy–Schwarz inequality becomes:

$${\displaystyle \langle \mathbf {u} ,\mathbf {v} \rangle ^{2}=(\|\mathbf {u} \|\|\mathbf {v} \|\cos \theta )^{2}\leq \|\mathbf {u} \|^{2}\|\mathbf {v} \|^{2},}$$

where ${\displaystyle \theta }$ is the angle between ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$.

The form above is perhaps the easiest in which to understand the inequality, since the square of the cosine can be at most 1, which occurs when the vectors are in the same or opposite directions. It can also be restated in terms of the vector coordinates ${\displaystyle u_{1}}$, ${\displaystyle u_{2}}$, ${\displaystyle v_{1}}$, and ${\displaystyle v_{2}}$ as

$${\displaystyle \left(u_{1}v_{1}+u_{2}v_{2}\right)^{2}\leq \left(u_{1}^{2}+u_{2}^{2}\right)\left(v_{1}^{2}+v_{2}^{2}\right),}$$

where equality holds if and only if the vector ${\displaystyle \left(u_{1},u_{2}\right)}$ is in the same or opposite direction as the vector ${\displaystyle \left(v_{1},v_{2}\right)}$, or if one of them is the zero vector.

In Euclidean space ${\displaystyle \mathbb {R} ^{n}}$ with the standard inner product, which is the dot product, the Cauchy–Schwarz inequality becomes:

$${\displaystyle \left(\sum _{i=1}^{n}u_{i}v_{i}\right)^{2}\leq \left(\sum _{i=1}^{n}u_{i}^{2}\right)\left(\sum _{i=1}^{n}v_{i}^{2}\right).}$$

The Cauchy–Schwarz inequality can be proved using only elementary algebra in this case by observing that the difference of the right and the left hand side is ${\displaystyle {\frac {1}{2}}\sum _{i,j=1,\ldots ,n}(u_{i}v_{j}-u_{j}v_{i})^{2}\geq 0}$

or by considering the following quadratic polynomial in ${\displaystyle x}$

$${\displaystyle \left(u_{1}x+v_{1}\right)^{2}+\cdots +\left(u_{n}x+v_{n}\right)^{2}=\left(\sum _{i}u_{i}^{2}\right)x^{2}+2\left(\sum _{i}u_{i}v_{i}\right)x+\sum _{i}v_{i}^{2}.}$$

Since the latter polynomial is nonnegative, it has at most one real root, hence its discriminant is less than or equal to zero. That is,

$${\displaystyle \left(\sum _{i}u_{i}v_{i}\right)^{2}-\left(\sum _{i}{u_{i}^{2}}\right)\left(\sum _{i}{v_{i}^{2}}\right)\leq 0.}$$

If ${\displaystyle \mathbf {u} ,\mathbf {v} \in \mathbb {C} ^{n}}$ with ${\displaystyle \mathbf {u} =\left(u_{1},\ldots ,u_{n}\right)}$ and ${\displaystyle \mathbf {v} =\left(v_{1},\ldots ,v_{n}\right)}$ (where ${\displaystyle u_{1},\ldots ,u_{n}\in \mathbb {C} }$ and ${\displaystyle v_{1},\ldots ,v_{n}\in \mathbb {C} }$) and if the inner product on the vector space ${\displaystyle \mathbb {C} ^{n}}$ is the canonical complex inner product (defined by ${\displaystyle \langle \mathbf {u} ,\mathbf {v} \rangle$ :=u_{1}{\overline {v_{1}}}+\cdots +u_{n}{\overline {v_{n}}},} where the bar notation is used for complex conjugation), then the inequality may be restated more explicitly as follows:

$${\displaystyle |\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}=\left|\sum _{k=1}^{n}u_{k}{\bar {v}}_{k}\right|^{2}\leq \langle \mathbf {u} ,\mathbf {u} \rangle \langle \mathbf {v} ,\mathbf {v} \rangle =\left(\sum _{k=1}^{n}u_{k}{\bar {u}}_{k}\right)\left(\sum _{k=1}^{n}v_{k}{\bar {v}}_{k}\right)=\sum _{j=1}^{n}\left|u_{j}\right|^{2}\sum _{k=1}^{n}\left|v_{k}\right|^{2}.}$$

That is,

$${\displaystyle \left|u_{1}{\bar {v}}_{1}+\cdots +u_{n}{\bar {v}}_{n}\right|^{2}\leq \left(\left|u_{1}\right|^{2}+\cdots +\left|u_{n}\right|^{2}\right)\left(\left|v_{1}\right|^{2}+\cdots +\left|v_{n}\right|^{2}\right).}$$

For the inner product space of square-integrable complex-valued functions, the following inequality:

$${\displaystyle \left|\int _{\mathbb {R} ^{n}}f(x){\overline {g(x)}}\,dx\right|^{2}\leq \int _{\mathbb {R} ^{n}}|f(x)|^{2}\,dx\int _{\mathbb {R} ^{n}}|g(x)|^{2}\,dx.}$$

The Hölder inequality is a generalization of this.

In any inner product space, the triangle inequality is a consequence of the Cauchy–Schwarz inequality, as is now shown:

$${\displaystyle {\begin{alignedat}{4}\|\mathbf {u} +\mathbf {v} \|^{2}&=\langle \mathbf {u} +\mathbf {v} ,\mathbf {u} +\mathbf {v} \rangle &&\\&=\|\mathbf {u} \|^{2}+\langle \mathbf {u} ,\mathbf {v} \rangle +\langle \mathbf {v} ,\mathbf {u} \rangle +\|\mathbf {v} \|^{2}~&&~{\text{ where }}\langle \mathbf {v} ,\mathbf {u} \rangle ={\overline {\langle \mathbf {u} ,\mathbf {v} \rangle }}\\&=\|\mathbf {u} \|^{2}+2\operatorname {Re} \langle \mathbf {u} ,\mathbf {v} \rangle +\|\mathbf {v} \|^{2}&&\\&\leq \|\mathbf {u} \|^{2}+2|\langle \mathbf {u} ,\mathbf {v} \rangle |+\|\mathbf {v} \|^{2}&&\\&\leq \|\mathbf {u} \|^{2}+2\|\mathbf {u} \|\|\mathbf {v} \|+\|\mathbf {v} \|^{2}~&&~{\text{ using CS}}\\&=(\|\mathbf {u} \|+\|\mathbf {v} \|)^{2}.&&\end{alignedat}}}$$

Taking square roots gives the triangle inequality:

$${\displaystyle \|\mathbf {u} +\mathbf {v} \|\leq \|\mathbf {u} \|+\|\mathbf {v} \|.}$$

The Cauchy–Schwarz inequality is used to prove that the inner product is a continuous function with respect to the topology induced by the inner product itself.[11][12]

The Cauchy–Schwarz inequality allows one to extend the notion of "angle between two vectors" to any real inner-product space by defining:[13][14]

$${\displaystyle \cos \theta _{\mathbf {u} \mathbf {v} }={\frac {\langle \mathbf {u} ,\mathbf {v} \rangle }{\|\mathbf {u} \|\|\mathbf {v} \|}}.}$$

The Cauchy–Schwarz inequality proves that this definition is sensible, by showing that the right-hand side lies in the interval [−1, 1] and justifies the notion that (real) Hilbert spaces are simply generalizations of the Euclidean space. It can also be used to define an angle in complex inner-product spaces, by taking the absolute value or the real part of the right-hand side,[15][16] as is done when extracting a metric from quantum fidelity.

Let ${\displaystyle X}$ and ${\displaystyle Y}$ be random variables, then the covariance inequality[17][18] is given by:

$${\displaystyle \operatorname {Var} (X)\geq {\frac {\operatorname {Cov} (X,Y)^{2}}{\operatorname {Var} (Y)}}.}$$

After defining an inner product on the set of random variables using the expectation of their product,

$${\displaystyle \langle X,Y\rangle$$ :=\operatorname {E} (XY),}

the Cauchy–Schwarz inequality becomes

$${\displaystyle |\operatorname {E} (XY)|^{2}\leq \operatorname {E} (X^{2})\operatorname {E} (Y^{2}).}$$

To prove the covariance inequality using the Cauchy–Schwarz inequality, let ${\displaystyle \mu =\operatorname {E} (X)}$ and ${\displaystyle \nu =\operatorname {E} (Y),}$ then

$${\displaystyle {\begin{aligned}|\operatorname {Cov} (X,Y)|^{2}&=|\operatorname {E} ((X-\mu )(Y-\nu ))|^{2}\\&=|\langle X-\mu ,Y-\nu \rangle |^{2}\\&\leq \langle X-\mu ,X-\mu \rangle \langle Y-\nu ,Y-\nu \rangle \\&=\operatorname {E} \left((X-\mu )^{2}\right)\operatorname {E} \left((Y-\nu )^{2}\right)\\&=\operatorname {Var} (X)\operatorname {Var} (Y),\end{aligned}}}$$

where ${\displaystyle \operatorname {Var} }$ denotes variance and ${\displaystyle \operatorname {Cov} }$ denotes covariance.

The special case of ${\displaystyle \mathbf {v} =\mathbf {0} }$ was proven above so it is henceforth assumed that ${\displaystyle \mathbf {v} \neq \mathbf {0} .}$ The Cauchy–Schwarz inequality (and the rest of the theorem) is an almost immediate corollary of the following equality:[21]

${\displaystyle {\frac {1}{\|\mathbf {v} \|^{2}}}\left\|\|\mathbf {v} \|^{2}\mathbf {u} -\langle \mathbf {u} ,\mathbf {v} \rangle \mathbf {v} \right\|^{2}~=~\|\mathbf {u} \|^{2}\|\mathbf {v} \|^{2}-|\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}}$

(Eq. 1)

Equality Eq. 1 is readily verified by elementarily expanding ${\displaystyle \left\|\|\mathbf {v} \|^{2}\mathbf {u} -\langle \mathbf {u} ,\mathbf {v} \rangle \mathbf {v} \right\|^{2}}$ (via the definition of the norm) and then simplifying:

Proof of Eq. 1

Let ${\displaystyle V=\|\mathbf {v} \|^{2}}$ and ${\displaystyle c=\langle \mathbf {u} ,\mathbf {v} \rangle }$ so that ${\displaystyle {\bar {c}}c=|c|^{2}=|\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}}$ and ${\displaystyle {\bar {c}}={\overline {\langle \mathbf {u} ,\mathbf {v} \rangle }}=\langle \mathbf {v} ,\mathbf {u} \rangle .}$ Then

$${\displaystyle {\begin{alignedat}{4}\left\|\|\mathbf {v} \|^{2}\mathbf {u} -\langle \mathbf {u} ,\mathbf {v} \rangle \mathbf {v} \right\|^{2}&=\|V\mathbf {u} -c\mathbf {v} \|^{2}=\langle V\mathbf {u} -c\mathbf {v} ,V\mathbf {u} -c\mathbf {v} \rangle &&~\quad {\text{ By definition of the norm }}\\[0.5ex]&=\langle V\mathbf {u} ,V\mathbf {u} \rangle -\langle V\mathbf {u} ,c\mathbf {v} \rangle -\langle c\mathbf {v} ,V\mathbf {u} \rangle +\langle c\mathbf {v} ,c\mathbf {v} \rangle &&~\quad {\text{ Expand }}\\[0.5ex]&=V^{2}\langle \mathbf {u} ,\mathbf {u} \rangle -V{\bar {c}}\langle \mathbf {u} ,\mathbf {v} \rangle -cV\langle \mathbf {v} ,\mathbf {u} \rangle \,+c{\bar {c}}\langle \mathbf {v} ,\mathbf {v} \rangle &&~\quad {\text{ Pull out scalars (note that }}V:=\|\mathbf {v} \|^{2}{\text{ is real) }}\\[0.5ex]&=V^{2}\|\mathbf {u} \|^{2}~~-V{\bar {c}}c~~~~~~~~-cV{\bar {c}}~~~~~~~~+c{\bar {c}}V&&~\quad {\text{ Use definitions of }}c:=\langle \mathbf {u} ,\mathbf {v} \rangle {\text{ and }}V\\[0.5ex]&=V^{2}\|\mathbf {u} \|^{2}~~-V{\bar {c}}c~=~V\left[V\|\mathbf {u} \|^{2}-{\bar {c}}c\right]&&~\quad {\text{ Simplify }}\\[0.5ex]&=\|\mathbf {v} \|^{2}\left[\|\mathbf {u} \|^{2}\|\mathbf {v} \|^{2}-|\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}\right]&&~\quad {\text{ Rewrite in terms of }}\mathbf {u} {\text{ and }}\mathbf {v} .\\\end{alignedat}}}$$

Dividing by ${\displaystyle \|\mathbf {v} \|^{2}\neq 0}$ completes the proof. ${\displaystyle \blacksquare }$

This expansion does not require ${\displaystyle \mathbf {v} }$ to be non-zero; however, ${\displaystyle \mathbf {v} }$ must be non-zero in order to divide both sides by ${\displaystyle \|\mathbf {v} \|^{2}}$ and to deduce the Cauchy–Schwarz inequality from it. Swapping ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$ gives rise to:

$${\displaystyle \left\|\|\mathbf {u} \|^{2}\mathbf {v} -{\overline {\langle \mathbf {u} ,\mathbf {v} \rangle }}\mathbf {u} \right\|^{2}~=~\|\mathbf {u} \|^{2}\left[\|\mathbf {u} \|^{2}\|\mathbf {v} \|^{2}-|\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}\right]}$$

and thus

$${\displaystyle {\begin{alignedat}{4}\|\mathbf {u} \|^{2}\|\mathbf {v} \|^{2}\left[\|\mathbf {u} \|^{2}\|\mathbf {v} \|^{2}-|\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}\right]~&=~\|\mathbf {u} \|^{2}\left\|\|\mathbf {v} \|^{2}\mathbf {u} -\langle \mathbf {u} ,\mathbf {v} \rangle \mathbf {v} \right\|^{2}\\~&=~\|\mathbf {v} \|^{2}\left\|\|\mathbf {u} \|^{2}\mathbf {v} -{\overline {\langle \mathbf {u} ,\mathbf {v} \rangle }}\mathbf {u} \right\|^{2}.\\\end{alignedat}}}$$

The special case of ${\displaystyle \mathbf {v} =\mathbf {0} }$ was proven above so it is henceforth assumed that ${\displaystyle \mathbf {v} \neq \mathbf {0} .}$ Let

$${\displaystyle \mathbf {z}$$ :=\mathbf {u} -{\frac {\langle \mathbf {u} ,\mathbf {v} \rangle }{\langle \mathbf {v} ,\mathbf {v} \rangle }}\mathbf {v} .}

It follows from the linearity of the inner product in its first argument that:

$${\displaystyle \langle \mathbf {z} ,\mathbf {v} \rangle =\left\langle \mathbf {u} -{\frac {\langle \mathbf {u} ,\mathbf {v} \rangle }{\langle \mathbf {v} ,\mathbf {v} \rangle }}\mathbf {v} ,\mathbf {v} \right\rangle =\langle \mathbf {u} ,\mathbf {v} \rangle -{\frac {\langle \mathbf {u} ,\mathbf {v} \rangle }{\langle \mathbf {v} ,\mathbf {v} \rangle }}\langle \mathbf {v} ,\mathbf {v} \rangle =0.}$$

Therefore, ${\displaystyle \mathbf {z} }$ is a vector orthogonal to the vector ${\displaystyle \mathbf {v} }$ (Indeed, ${\displaystyle \mathbf {z} }$ is the projection of ${\displaystyle \mathbf {u} }$ onto the plane orthogonal to ${\displaystyle \mathbf {v} .}$) We can thus apply the Pythagorean theorem to

$${\displaystyle \mathbf {u} ={\frac {\langle \mathbf {u} ,\mathbf {v} \rangle }{\langle \mathbf {v} ,\mathbf {v} \rangle }}\mathbf {v} +\mathbf {z} }$$

which gives

$${\displaystyle \|\mathbf {u} \|^{2}=\left|{\frac {\langle \mathbf {u} ,\mathbf {v} \rangle }{\langle \mathbf {v} ,\mathbf {v} \rangle }}\right|^{2}\|\mathbf {v} \|^{2}+\|\mathbf {z} \|^{2}={\frac {|\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}}{(\|\mathbf {v} \|^{2})^{2}}}\,\|\mathbf {v} \|^{2}+\|\mathbf {z} \|^{2}={\frac {|\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}}{\|\mathbf {v} \|^{2}}}+\|\mathbf {z} \|^{2}\geq {\frac {|\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}}{\|\mathbf {v} \|^{2}}}.}$$

The Cauchy–Schwarz inequality follows by multiplying by ${\displaystyle \|\mathbf {v} \|^{2}}$ and then taking the square root. Moreover, if the relation ${\displaystyle \geq }$ in the above expression is actually an equality, then ${\displaystyle \|\mathbf {z} \|^{2}=0}$ and hence ${\displaystyle \mathbf {z} =\mathbf {0}$ ;} the definition of ${\displaystyle \mathbf {z} }$ then establishes a relation of linear dependence between ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} .}$ The converse was proved at the beginning of this section, so the proof is complete. ${\displaystyle \blacksquare }$

Let ${\displaystyle (V,\langle \cdot ,\cdot \rangle )}$ be a real inner product space. Consider an arbitrary pair ${\displaystyle \mathbf {u} ,\mathbf {v} \in V}$ and the function ${\displaystyle p:\mathbb {R} \to \mathbb {R} }$ defined by ${\displaystyle p(t)=\langle t\mathbf {u} +\mathbf {v} ,t\mathbf {u} +\mathbf {v} \rangle .}$ Since the inner product is positive-definite, ${\displaystyle p(t)}$ only takes non-negative values. On the other hand, ${\displaystyle p(t)}$ can be expanded using the bilinearity of the inner product and using the fact that ${\displaystyle \langle \mathbf {u} ,\mathbf {v} \rangle =\langle \mathbf {v} ,\mathbf {u} \rangle }$ for real inner products:

$${\displaystyle p(t)=\Vert \mathbf {u} \Vert ^{2}t^{2}+t\left[\langle \mathbf {u} ,\mathbf {v} \rangle +\langle \mathbf {v} ,\mathbf {u} \rangle \right]+\Vert \mathbf {v} \Vert ^{2}=\Vert \mathbf {u} \Vert ^{2}t^{2}+2t\langle \mathbf {u} ,\mathbf {v} \rangle +\Vert \mathbf {v} \Vert ^{2}.}$$

Thus, ${\displaystyle p}$ is a polynomial of degree ${\displaystyle 2}$ (unless ${\displaystyle u=0,}$ which is a case that can be independently verified). Since the sign of ${\displaystyle p}$ does not change, the discriminant of this polynomial must be non-positive:

$${\displaystyle \Delta =4\left(\langle \mathbf {u} ,\mathbf {v} \rangle ^{2}-\Vert \mathbf {u} \Vert ^{2}\Vert \mathbf {v} \Vert ^{2}\right)\leqslant 0.}$$

The conclusion follows.

For the equality case, notice that ${\displaystyle \Delta =0}$ happens if and only if ${\displaystyle p(t)=(t\Vert \mathbf {u} \Vert +\Vert \mathbf {v} \Vert )^{2}.}$ If ${\displaystyle t_{0}=-\Vert \mathbf {v} \Vert /\Vert \mathbf {u} \Vert ,}$ then ${\displaystyle p(t_{0})=\langle t_{0}\mathbf {u} +\mathbf {v} ,t_{0}\mathbf {u} +\mathbf {v} \rangle =0,}$ and hence ${\displaystyle \mathbf {v} =-t_{0}\mathbf {u} .}$

The Cauchy–Schwarz inequality in the case where the inner product is the dot product on ${\displaystyle \mathbb {R} ^{n}}$ is now proven. The Cauchy–Schwarz inequality may be rewritten as ${\displaystyle \left|\mathbf {a} \cdot \mathbf {b} \right|^{2}\leq \left\|\mathbf {a} \right\|^{2}\,\left\|\mathbf {b} \right\|^{2}}$ or equivalently, ${\displaystyle \left(\mathbf {a} \cdot \mathbf {b} \right)^{2}\leq \left(\mathbf {a} \cdot \mathbf {a} \right)\,\left(\mathbf {b} \cdot \mathbf {b} \right)}$ for ${\displaystyle \mathbf {a}$ :=\left(a_{1},\ldots ,a_{n}\right),\mathbf {b} :=\left(b_{1},\ldots ,b_{n}\right)\in \mathbb {R} ^{n},} which expands to:

$${\displaystyle \left(a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+\cdots +b_{n}^{2}\right)\geq \left(a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}\right)^{2}.}$$

To simplify, let

$${\displaystyle {\begin{aligned}A&=a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2},\\B&=b_{1}^{2}+b_{2}^{2}+\cdots +b_{n}^{2}\\D&=a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}\\\end{aligned}}}$$

so that the statement that remains to be to proven can be written as ${\displaystyle AB\geq D^{2},}$ which can be rearranged to ${\displaystyle D^{2}-AB\leq 0.}$ The discriminant of the quadratic equation ${\displaystyle Ax^{2}+2Dx+B}$ is ${\displaystyle 4D^{2}-4AB.}$

Therefore, to complete the proof it is sufficient to prove that this quadratic either has no real roots or exactly one real root, because this will imply:

$${\displaystyle 4\left(D^{2}-AB\right)\leq 0.}$$

Substituting the values of ${\displaystyle A,B,D}$ into ${\displaystyle Ax^{2}+2Dx+B}$ gives:

$${\displaystyle {\begin{alignedat}{4}Ax^{2}+2Dx+B&=\left(a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}\right)x^{2}+2\left(a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}\right)x+\left(b_{1}^{2}+b_{2}^{2}+\cdots +b_{n}^{2}\right)\\&=\left(a_{1}^{2}x^{2}+2a_{1}b_{1}x+b_{1}^{2}\right)+\left(a_{2}^{2}x^{2}+2a_{2}b_{2}x+b_{2}^{2}\right)+\cdots +\left(a_{n}^{2}x^{2}+2a_{n}b_{n}x+b_{n}^{2}\right)\\&=\left(a_{1}x+b_{1}\right)^{2}+\left(a_{2}x+b_{2}\right)^{2}+\cdots +\left(a_{n}x+b_{n}\right)^{2}\\&\geq 0\end{alignedat}}}$$

which is a sum of terms that are each ${\displaystyle \,\geq 0\,}$ by the trivial inequality: ${\displaystyle r^{2}\geq 0}$ for all ${\displaystyle r\in \mathbb {R} .}$ This proves the inequality and so to finish the proof, it remains to show that equality is achievable. The equality ${\displaystyle a_{i}x=-b_{i}}$ is the equality case for Cauchy–Schwarz after inspecting

$${\displaystyle \left(a_{1}x+b_{1}\right)^{2}+\left(a_{2}x+b_{2}\right)^{2}+\cdots +\left(a_{n}x+b_{n}\right)^{2}\geq 0,}$$

which proves that equality is achievable. ${\displaystyle \blacksquare }$