Chebyshev’s inequality

Proof

Integrate this inequality with respect to x {displaystyle x} and y {displaystyle y} over [ a , b ] {displaystyle [a,b]} : ∫ a b ∫ a b ( f ( x ) − f ( y ) ) ( g ( x ) − g ( y ) ) d x d y ≥ 0. {displaystyle int _{a}^{b}int _{a}^{b}(f(x)-f(y))(g(x)-g(y)),dx,dygeq 0.}

Expanding the integrand gives: ∫ a b ∫ a b [ f ( x ) g ( x ) − f ( x ) g ( y ) − f ( y ) g ( x ) + f ( y ) g ( y ) ] d x d y ≥ 0. {displaystyle int _{a}^{b}int _{a}^{b}left[f(x)g(x)-f(x)g(y)-f(y)g(x)+f(y)g(y)right],dx,dygeq 0.}

Separate the double integral into four parts: ∫ a b ∫ a b f ( x ) g ( x ) d x d y − ∫ a b ∫ a b f ( x ) g ( y ) d x d y − ∫ a b ∫ a b f ( y ) g ( x ) d x d y + ∫ a b ∫ a b f ( y ) g ( y ) d x d y ≥ 0. {displaystyle int _{a}^{b}int _{a}^{b}f(x)g(x),dx,dy-int _{a}^{b}int _{a}^{b}f(x)g(y),dx,dy-int _{a}^{b}int _{a}^{b}f(y)g(x),dx,dy+int _{a}^{b}int _{a}^{b}f(y)g(y),dx,dygeq 0.}

Since the integration variable in each inner integral is independent, we have:

• ∫ a b ∫ a b f ( x ) g ( x ) d x d y = ( b − a ) ∫ a b f ( x ) g ( x ) d x , {displaystyle int _{a}^{b}int _{a}^{b}f(x)g(x),dx,dy=(b-a)int _{a}^{b}f(x)g(x),dx,}
• ∫ a b ∫ a b f ( y ) g ( y ) d x d y = ( b − a ) ∫ a b f ( y ) g ( y ) d y = ( b − a ) ∫ a b f ( x ) g ( x ) d x , {displaystyle int _{a}^{b}int _{a}^{b}f(y)g(y),dx,dy=(b-a)int _{a}^{b}f(y)g(y),dy=(b-a)int _{a}^{b}f(x)g(x),dx,}
• ∫ a b ∫ a b f ( x ) g ( y ) d x d y = ( ∫ a b f ( x ) d x ) ( ∫ a b g ( y ) d y ) , {displaystyle int _{a}^{b}int _{a}^{b}f(x)g(y),dx,dy=left(int _{a}^{b}f(x),dxright)left(int _{a}^{b}g(y),dyright),}
• ∫ a b ∫ a b f ( y ) g ( x ) d x d y = ( ∫ a b f ( y ) d y ) ( ∫ a b g ( x ) d x ) = ( ∫ a b f ( x ) d x ) ( ∫ a b g ( x ) d x ) . {displaystyle int _{a}^{b}int _{a}^{b}f(y)g(x),dx,dy=left(int _{a}^{b}f(y),dyright)left(int _{a}^{b}g(x),dxright)=left(int _{a}^{b}f(x),dxright)left(int _{a}^{b}g(x),dxright).}

Let

I = ∫ a b f ( x ) g ( x ) d x , F = ∫ a b f ( x ) d x , G = ∫ a b g ( x ) d x . {displaystyle I=int _{a}^{b}f(x)g(x),dx,quad F=int _{a}^{b}f(x),dx,quad G=int _{a}^{b}g(x),dx.}

Substitute these into the inequality:

( b − a ) I − F G − F G + ( b − a ) I ≥ 0. {displaystyle (b-a)I-FG-FG+(b-a)Igeq 0.}

Simplify:

2 ( b − a ) I − 2 F G ≥ 0. {displaystyle 2(b-a)I-2FGgeq 0.}

Dividing by 2 ( b − a ) {displaystyle 2(b-a)} (noting that b − a > 0 {displaystyle b-a>0} ):

I ≥ F G ( b − a ) . {displaystyle Igeq {frac {FG}{(b-a)}}.}

Divide both sides by b − a {displaystyle b-a} to obtain:

1 b − a ∫ a b f ( x ) g ( x ) d x ≥ ( 1 b − a ∫ a b f ( x ) d x ) ( 1 b − a ∫ a b g ( x ) d x ) . {displaystyle {frac {1}{b-a}}int _{a}^{b}f(x)g(x),dxgeq left({frac {1}{b-a}}int _{a}^{b}f(x),dxright)left({frac {1}{b-a}}int _{a}^{b}g(x),dxright).}

This completes the proof.