Cauchy’s inequality:If $x_i,y_i(1\leq i\leq n)$ are non-negative real numbers,then
$$(x_1y_1+\cdots+x_ny_n)^2\leq (x_1^2+\cdots+x_n^2)(y_1^2+\dots+y_n^2)$$
The equality holds if and only if $(x_1,\cdots ,x_n)$ and ${(y_1,\cdots ,y_n)}$ are linearly dependent.
Proof:When ${n=1}$,the inequality obviously holds.And it is easy to verify that ${x_1}$ and ${y_1}$ are linearly dependent.Suppose when ${n=k}$,the inequality holds,that is
$$(x_1y_1+\cdots +x_ky_k)^2\leq (x_1^2+\cdots +x_k^2)(y_1^2+\cdots +y_k^2)$$
And the equality holds if and only if ${(x_1,\cdots,x_k)}$ and ${(y_1,\cdots,y_k)}$ are linearly dependent.In case of ${n=k+1}$,
\begin{align*}(x_1y_1+\cdots +x_ky_k+x_{k+1}y_{k+1})^2&=(x_1y_1+\cdots +x_ky_k)^2+2x_{k+1}y_{k+1}(x_1y_1+\cdots+ x_ky_k)+(x_{k+1}y_{k+1})^2\\&\leq (x_1^2+\cdots +x_k^2)(y_1^2+\cdots +y_k^2)+2x_{k+1}y_{k+1}(x_1y_1+\cdots +x_ky_k)+(x_{k+1}y_{k+1})^2\\&\leq (x_1^2+\cdots +x_k^2)(y_1^2+\cdots +y_k^2)+2x_{k+1}y_{k+1}\sqrt{(x_1^2 +\cdots +x_k^2)(y_1^2+\cdots +y_k^2)}+(x_{k+1}y_{k+1})^2\\&\leq (x_1^2+\cdots +x_k^2)(y_1^2+\cdots +y_k^2)+x_{k+1}^2(y_1^2+\cdots +y_k^2)+y_{k+1}^2(x_1^2+\cdots +x_k^2)+(x_{k+1}y_{k+1})^2\\&=(x_1^2+\cdots +x_{k+1}^2)(y_1^2+\cdots +y_{k+1}^2)\end{align*}
The equality holds if and only if :
- ${(x_1,\cdots,x_k)}$ and ${(y_1,\cdots,y_k)}$ are linearly dependent.
- ${x_{k+1}\sqrt{y_1^2+\cdots +y_k^2}=y_{k+1}\sqrt{x_1^2+\cdots +x_k^2}}$
${(x_1,\cdots ,x_k)}$ and ${(y_1,\cdots ,y_k)}$ are linearly dependent,which means that ${a_1(x_1,\cdots ,x_k)+a_2(y_1,\cdots,y_k)=0}$ while ${a_1}$ or ${a_2}$ are not ${0}$,if ${a_1=0}$,then all of ${y_i(1\leq i\leq k)}$ are 0,so ${y_{k+1}=0}$ or all of ${x_i(1\leq i\leq k)}$ are 0,whatever the case,we can easily verify that ${(x_1,\cdots ,x_{k+1})}$ and ${(y_1,\cdots ,y_{k+1})}$ are linearly dependent.Similary,when ${a_2=0}$ ,we can get the same conclusion.When both ${a_1}$ and ${a_2}$ are not 0,then $${(x_1,\cdots,x_k)=\lambda (y_1,\cdots,y_k)(\lambda\neq 0)}$$so $${\frac{\sqrt{x_1^2+\cdots +x_k^2}}{\sqrt{y_1^2+\cdots +y_k^2}}=\lambda}$$ so $${\frac{x_{k+1}}{y_{k+1}}=\lambda }$$so ${(x_1,\cdots,x_{k+1})}$ and ${(y_1,\cdots,y_{k+1})}$ are linearly dependent.
And,it is easy to verify that when ${(x_1,\cdots,x_n)}$ and ${(y_1,\cdots,y_n)}$ are linearly dependent,the equality holds.${\Box}$
Now I’d like to talk about another point.Why ${x_i,y_i(1\leq i\leq n)}$ should be non-negative?In fact,this prerequisite is not necessary,because
$$|x_1y_1+\cdots +x_ny_n|\leq |x_1||y_1|+\cdots +|x_n||y_n|$$
The equality holds if and only if ${x_1y_1,\cdots,x_ny_n}$ are all non-negative or all negative.So,$${(x_1y_1+\cdots+x_ny_n)^2\leq (|x_1||y_1|+\cdots +|x_n||y_n|)^2\leq (x_1^2+\cdots+x_n^2)(y_1^2+\dots+y_n^2)}$$
The equality holds iff :
- ${(|x_1|,\cdots,|x_n|)}$ and ${(|y_1|,\cdots,|y_n|)}$ are linearly dependent.
- ${x_1y_1,\cdots,x_ny_n}$ are all non-negative or all negative .
It is easy to verify that the above two point is equivalent to this point: ${(x_1,\cdots,x_n)}$ and ${(y_1,\cdots,y_n)}$ are linearly dependent.