Question

Demonstrati inegalitatea:
$\frac{b-a}{1+ b^{2} } \ \textless \ arctg \frac{b-a}{1+ab} \ \textless \ \frac{b-a}{1+ a^{2} }$ , 0<a<b

Answer 1

[tex]\dfrac{b-a}{1+a^2}\ \textless \ arctg \dfrac{b-a}{1+ab}\ \textless \ \dfrac{b-a}{1+b^2},0\ \textless \ a\ \textless \ b\\ \text{Impartind la b-a se obtine:}\\ \dfrac{1}{1+a^2}\ \textless \ \dfrac{arctg \frac{b-a}{1+ab}}{b-a}\ \textless \ \dfrac{1}{1+b^2}\\ \dfrac{1}{1+a^2}\ \textless \ \dfrac{arctg\ b-arctg\ a}{b-a}\ \textless \ \dfrac{1}{1+b^2}\\ \text{Consideram functia:}\\ f:[a,b]\Rightarrow \mathbb{R},f(x)=arctg x\\ \text{Functia este continua si derivabila,deci conform teoremei lui Lagrange}\\ \exists c\in (a,b)\text{ pentru care }\\ f'(c)=\dfrac{f(b)-f(a)}{b-a}=\dfrac{arctg\ b-arctg\ a}{b-a} [/tex][tex]\text{Dar }f'(c)=\dfrac{1}{1+c^2}\\ \text{Inegalitatea se reduce la :}\\ \dfrac{1}{1+b^2}\ \textless \ \dfrac{1}{1+c^2}\ \textless \ \dfrac{1}{1+a^2}|()^{-1}\\ 1+b^2\ \textgreater \ 1+c^2\ \textgreater \ 1+a^2|-1\\ b^2\ \textgreater \ c^2\ \textgreater \ a^2 |\sqrt{} \\ b\ \textgreater \ c\ \textgreater \ a,\text{ceea ce este adevarat.}[/tex]