Question

Cho a,b>0 và ab+a+b=1 . Chứng minh:

\(\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}=\dfrac{1+ab}{\sqrt{2\left(1+a^2\right)\left(1+b^2\right)}}\)

Answer 1

Lời giải:

Từ \(ab+a+b=1\) suy ra :

\(a^2+1=a^2+ab+a+b=a(a+b)+(a+b)=(a+1)(a+b)\)

\(b^2+1=b^2+ab+a+b=b(b+a)+(a+b)=(b+1)(a+b)\)

Do đó:
\(\frac{a}{a^2+1}+\frac{b}{b^2+1}=\frac{a}{(a+1)(a+b)}+\frac{b}{(b+1)(a+b)}=\frac{a(b+1)+b(a+1)}{(a+1)(b+1)(a+b)}\)

\(=\frac{ab+(ab+a+b)}{(a+1)(b+1)(a+b)}=\frac{ab+1}{(a+1)(b+1)(a+b)}(*)\)

Và:

\(\frac{ab+1}{\sqrt{2(a^2+1)(b^2+1)}}=\frac{ab+1}{\sqrt{2(a+1)(a+b)(b+1)(a+b)}}=\frac{ab+1}{\sqrt{(ab+a+b+1)(a+1)(b+1)(a+b)^2}}\)

\(=\frac{ab+1}{\sqrt{(a+1)(b+1)(a+1)(b+1)(a+b)^2}}=\frac{ab+1}{(a+1)(b+1)(a+b)}(**)\)

Từ $(*); (**)$ ta có đpcm.