Exercise:
Let $a,b\in \mathbb{R}$ such that , $a,b>1$
Show that, $\int_{-1}^{1}{\left( \frac{1+{{b}^{\left| x \right|}}}{1+{{a}^{x}}}+\frac{1+{{a}^{\left| x \right|}}}{1+{{b}^{x}}} \right)dx}<a+b+2$
Solution: we know that $\left| x \right|=0\Leftrightarrow x=0$
So $A=\int_{-1}^{1}{\left( \frac{1+{{b}^{\left| x \right|}}}{1+{{a}^{x}}}+\frac{1+{{a}^{\left| x \right|}}}{1+{{b}^{x}}} \right)dx}=\int_{-1}^{0}{\left( \frac{1+{{b}^{\left| x \right|}}}{1+{{a}^{x}}}+\frac{1+{{a}^{\left| x \right|}}}{1+{{b}^{x}}} \right)dx+\int_{0}^{1}{\left( \frac{1+{{b}^{\left| x \right|}}}{1+{{a}^{x}}}+\frac{1+{{a}^{\left| x \right|}}}{1+{{b}^{x}}} \right)dx}}$
$=\int_{-1}^{0}{\left( \frac{1+{{b}^{-x}}}{1+{{a}^{x}}}+\frac{1+{{a}^{-x}}}{1+{{b}^{x}}} \right)dx+\int_{0}^{1}{\left( \frac{1+{{b}^{x}}}{1+{{a}^{x}}}+\frac{1+{{a}^{x}}}{1+{{b}^{x}}} \right)dx}}$
Let ${{A}_{1}}=\int_{-1}^{0}{\left( \frac{1+{{b}^{-x}}}{1+{{a}^{x}}}+\frac{1+{{a}^{-x}}}{1+{{b}^{x}}} \right)dx}$
Put $u=x+1\Leftrightarrow du=dx\,\,\And \,\,x=u-1$
${{A}_{1}}=\int_{u(-1)}^{u(0)}{\left( \frac{1+{{b}^{-\left( u-1 \right)}}}{1+{{a}^{u-1}}}+\frac{1+{{a}^{-\left( u-1 \right)}}}{1+{{b}^{u-1}}} \right)du}=\int_{0}^{1}{\left( \frac{1+{{b}^{1-u}}}{1+{{a}^{u-1}}}+\frac{1+{{a}^{1-u}}}{1+{{b}^{u-1}}} \right)du}$
$=\int_{0}^{1}{\left( \frac{1+{{b}^{1-u}}}{1+{{a}^{-\left( 1-u \right)}}}+\frac{1+{{a}^{1-u}}}{1+{{b}^{-\left( 1-u \right)}}} \right)du}=\int_{0}^{1}{\left( \frac{1+{{b}^{x}}}{1+{{a}^{-x}}}+\frac{1+{{a}^{x}}}{1+{{b}^{-x}}} \right)dx}$
But $\frac{1+{{b}^{x}}}{1+\frac{1}{{{a}^{x}}}}=\frac{1+{{b}^{x}}}{\frac{{{a}^{x}}+1}{{{a}^{x}}}}=\frac{{{a}^{x}}\left( 1+{{b}^{x}} \right)}{{{a}^{x}}+1}\,\,\,\,\,\And \,\,\,\,\,\,\,\frac{1+{{a}^{x}}}{1+{{b}^{-x}}}=\frac{{{b}^{x}}\left( 1+{{a}^{x}} \right)}{{{b}^{x}}+1}$
Hence ${{A}_{1}}=\int_{0}^{1}{\left( \frac{{{a}^{x}}\left( 1+{{b}^{x}} \right)}{1+{{a}^{x}}}+\frac{{{b}^{x}}\left( 1+{{a}^{x}} \right)}{1+{{b}^{x}}} \right)dx}$
So $A=\int_{0}^{1}{\left( \frac{{{a}^{x}}\left( 1+{{b}^{x}} \right)}{1+{{a}^{x}}}+\frac{1+{{b}^{x}}}{1+{{a}^{x}}}+\frac{{{b}^{x}}\left( 1+{{a}^{x}} \right)}{1+{{b}^{x}}}+\frac{1+{{a}^{x}}}{1+{{b}^{x}}} \right)dx}$
$=\int_{0}^{1}{\left( \frac{\left( 1+{{a}^{x}} \right)\left( 1+{{b}^{x}} \right)}{1+{{a}^{x}}}+\frac{\left( 1+{{b}^{x}} \right)\left( 1+{{a}^{x}} \right)}{1+{{b}^{x}}} \right)dx}$
$=\int_{0}^{1}{\frac{\left( 1+{{a}^{x}} \right){{\left( 1+{{b}^{x}} \right)}^{2}}+{{\left( 1+{{a}^{x}} \right)}^{2}}\left( 1+{{b}^{x}} \right)}{\left( 1+{{a}^{x}} \right)\left( 1+{{b}^{x}} \right)}dx}$
$=\int_{0}^{1}{\frac{\left( 1+{{a}^{x}} \right)\left( 1+{{b}^{x}} \right)\left[ 1+{{b}^{x}}+1+{{a}^{x}} \right]}{\left( 1+{{a}^{x}} \right)\left( 1+{{b}^{x}} \right)}dx}$
$=\int_{0}^{1}{\left( {{a}^{x}}+{{b}^{x}}+2 \right)dx}=\int_{0}^{1}{{{a}^{x}}dx}+\int_{0}^{1}{{{b}^{x}}dx+2}\int_{0}^{1}{dx}$
Let $v={{a}^{x}}\Leftrightarrow dv={{a}^{x}}\ln a\,dx\Leftrightarrow dv=v\ln adx$
So $\int_{0}^{1}{{{a}^{x}}dx}=\int_{1}^{a}{\frac{v}{v\ln a}dv}=\int_{1}^{a}{\frac{1}{\ln a}dv}=\frac{a-1}{\ln a}$
Hence $A=\frac{a-1}{\ln a}+\frac{b-1}{\ln b}+2$
Now require to prove that $A<a+b+2$
Define a function $f:\left( 1,\infty \right)\to \mathbb{R}$ to be $f\left( x \right)=\frac{1}{x}+\ln x\,\,\,\,,x>1$ then $f\left( 1 \right)=1$
Then $f'\left( x \right)=-\frac{1}{{{x}^{2}}}+\frac{1}{x}=\frac{-1+x}{{{x}^{2}}}$ as $x>1\Leftrightarrow x-1>0$ hence $f'>0$
Thus $f$ is increasing function so $f\left( 1 \right)<f\left( x \right)$
That is $1<\frac{1}{a}+\ln a\,\,\,\,\,\And \,\,\,1<\frac{1}{b}+\ln b\Rightarrow 1-\frac{1}{a}<\ln a\,\,\And \,\,1-\frac{1}{b}<\ln a$
$\Rightarrow \frac{1}{\ln a}<\frac{a}{a-1}\,\,\,\And \,\,\,\,\frac{1}{\ln b}<\frac{b}{b-1}\Leftrightarrow \frac{a-1}{\ln a}<a\,\,\,\And \,\,\,\,\frac{b-1}{\ln b}<b$
Thus $\frac{a-1}{\ln a}+\frac{b-1}{\ln b}<a+b\Leftrightarrow A<a+b+2$ Q.E.D
*___________________
the idea of the solution credit to Hung Nguyen Viet
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