Prove Cauchy's inequality by induction
Cauchy’s inequality:If xi,yi(1≤i≤n) are non-negative real numbers,then
(x1y1+⋯+xnyn)2≤(x21+⋯+x2n)(y21+⋯+y2n)
The equality holds if and only if (x1,⋯,xn) and (y1,⋯,yn) are linearly dependent.
Proof:When n=1,the inequality obviously holds.And it is easy to verify that x1 and y1 are linearly dependent.Suppose when n=k,the inequality holds,that is
(x1y1+⋯+xkyk)2≤(x21+⋯+x2k)(y21+⋯+y2k)
And the equality holds if and only if (x1,⋯,xk) and (y1,⋯,yk) are linearly dependent.In case of n=k+1,
(x1y1+⋯+xkyk+xk+1yk+1)2=(x1y1+⋯+xkyk)2+2xk+1yk+1(x1y1+⋯+xkyk)+(xk+1yk+1)2≤(x21+⋯+x2k)(y21+⋯+y2k)+2xk+1yk+1(x1y1+⋯+xkyk)+(xk+1yk+1)2≤(x21+⋯+x2k)(y21+⋯+y2k)+2xk+1yk+1√(x21+⋯+x2k)(y21+⋯+y2k)+(xk+1yk+1)2≤(x21+⋯+x2k)(y21+⋯+y2k)+x2k+1(y21+⋯+y2k)+y2k+1(x21+⋯+x2k)+(xk+1yk+1)2=(x21+⋯+x2k+1)(y21+⋯+y2k+1)
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.
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