Integral inequality exercise asked by Dan Sitaru in the mathematical inequality group

Exercise:

Let $f\in {{C}^{1}}\left[ a,b \right]$ and let $m=\underset{a\le x\le b}{\mathop{\inf }}\,f'\left( x \right)\,\,\,\And \,\,\,M=\underset{a\le x\le b}{\mathop{\sup }}\,f'\left( x \right)$

Show that,

              $\frac{bf\left( b \right)-af\left( a \right)}{b-a}-\frac{M\left( a+b \right)}{2}\le \frac{1}{b-a}\int_{a}^{b}{f\left( x \right)dx\le }\frac{bf\left( b \right)-af\left( a \right)}{b-a}-\frac{m\left( a+b \right)}{2}$

Solution: Let $h\left( x \right)=xf\left( x \right)$ be a continuous function on $\left[ a,b \right]$ and differentiable on $\left( a,b \right)$

then by MVT there exists a real number $c$ such that $a\le c\le b$ such that $h'\left( c \right)=\frac{h\left( b \right)-h\left( a \right)}{b-a}$

but $h'\left( c \right)=f\left( c \right)+cf'\left( c \right)$ thus $f\left( c \right)+cf'\left( c \right)=\frac{bf\left( b \right)-af\left( a \right)}{b-a}$

We have $M=\underset{a\le x\le b}{\mathop{\sup }}\,f'\left( x \right)\Leftrightarrow f'\left( c \right)\le M\,\,\forall \,\,c\le b$ so $cf'\left( c \right)\le cM$

Thus $\frac{bf\left( b \right)-af\left( a \right)}{b-a}\le f\left( c \right)+cM\Leftrightarrow \frac{bf\left( b \right)-af\left( a \right)}{b-a}\int_{a}^{b}{dx}\le \int_{a}^{b}{f\left( x \right)dx}+M\int_{a}^{b}{x}\,dx$

$\Leftrightarrow \frac{bf\left( b \right)-af\left( a \right)}{b-a}\left( b-a \right)\le \int_{a}^{b}{f\left( x \right)dx+M\left( \frac{{{b}^{2}}-{{a}^{2}}}{2} \right)}$

$\Leftrightarrow \frac{bf\left( b \right)-af\left( a \right)}{b-a}\left( b-a \right)\le \int_{a}^{b}{f\left( x \right)dx}+M\left( \frac{\left( b-a \right)\left( b+a \right)}{2} \right)$

$\Leftrightarrow \frac{bf\left( b \right)-af\left( a \right)}{b-a}\le \frac{1}{b-a}\int_{a}^{b}{f\left( x \right)dx}+\frac{M\left( a+b \right)}{2}$                              (*)

Also we have $m=\underset{a\le x\le b}{\mathop{\inf }}\,f'\left( x \right)\Leftrightarrow m\le f'\left( c \right)\,\,\forall \,c\ge a\Leftrightarrow cm\le cf'\left( c \right)$

So $f\left( c \right)+cm\le \frac{bf\left( b \right)-af\left( a \right)}{b-a}\Leftrightarrow \int_{a}^{b}{f\left( x \right)dx+m\int_{a}^{b}{xdx}}\le \frac{bf\left( b \right)-af\left( a \right)}{b-a}\int_{a}^{b}{dx}$

$\Leftrightarrow \int_{a}^{b}{f\left( x \right)dx}+m\left( \frac{{{b}^{2}}-{{a}^{2}}}{2} \right)\le \frac{bf\left( b \right)-af\left( a \right)}{b-a}\left( b-a \right)$

$\Leftrightarrow \frac{1}{b-a}\int_{a}^{b}{f\left( x \right)dx+\frac{m\left( a+b \right)}{2}}\le \frac{bf\left( b \right)-af\left( a \right)}{b-a}$                              (**)

By (*) and (**) the proof is completed.                                    Q.E.D

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