Inequality Exercise Proved by using the MVT used to prove the holder inequality in Real analysis

Exercise:

Show that, ${{a}^{t}}{{b}^{1-t}}<at+b\left( 1-t \right)$ for $t\in \left( 0,1 \right)\,\,\And \,a,b\in {{\mathbb{R}}^{+}}$

Solution: let $f\left( x \right)={{x}^{t}}\,\,\,,\,\,\,0<t<1$ and $x>0$

Notice that $f$ is continuous in $\left[ a,b \right]$ and differentiable on $\left( a,b \right)$ , then

By MVT $\exists c\in \left[ a,b \right]$ such that $f'\left( c \right)=\frac{f\left( b \right)-f\left( a \right)}{b-a}=\frac{{{b}^{t}}-{{a}^{t}}}{b-a}$

We know that $f'\left( x \right)=t{{x}^{t-1}}\Leftrightarrow f'\left( c \right)=t{{c}^{t-1}}$ so $\frac{{{b}^{t}}-{{a}^{t}}}{b-a}=t{{c}^{t-1}}$

We have $c<b\Leftrightarrow {{b}^{-1}}<{{c}^{-1}}\Leftrightarrow {{b}^{t-1}}<{{c}^{t-1}}\Leftrightarrow t{{b}^{t-1}}<t{{c}^{t-1}}$

So $t{{b}^{t-1}}<\frac{{{b}^{t}}-{{a}^{t}}}{b-a}\Leftrightarrow \left( b-a \right)t{{b}^{t-1}}<{{b}^{t}}-{{a}^{t}}\Leftrightarrow bt{{b}^{t-1}}-at{{b}^{t-1}}<{{b}^{t}}-{{a}^{t}}$

$\Leftrightarrow t{{b}^{t}}<at{{b}^{t-1}}+{{b}^{t}}-{{a}^{t}}\Leftrightarrow {{a}^{t}}<at{{b}^{t-1}}+{{b}^{t}}-t{{b}^{t}}$ Multiply by ${{b}^{1-t}}$

$\Leftrightarrow {{a}^{t}}{{b}^{1-t}}<{{b}^{1-t}}\left( at{{b}^{t-1}}+{{b}^{t}}-t{{b}^{t}} \right)=at+{{b}^{1-t+t}}-t{{b}^{t+1-t}}=at+b-tb=at+b\left( 1-t \right)$

This Inequality used to prove the Holder Inequality In Real Analysis .