Question

[Bui Tien Hai Dang]

Answer 1

[Xyz OLM]

\(VT=\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}=\dfrac{a}{ab+a+b+a^2}+\dfrac{b}{ab+a+b+b^2}\)

\(=\dfrac{a}{\left(a+b\right).\left(a+1\right)}+\dfrac{b}{\left(a+b\right).\left(b+1\right)}\)

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

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

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

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

Từ (1) (2) => ĐPCM

Answer 2

[Nguyễn Lừ Thảm]

Giải

Với a,b > 0, ta có:

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

Tương đương

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

Mặt khác, \(\left(1+a^2\right)\left(1+b^2\right)=\left(a^2+a+b+ab\right)\left(b^2+a+b+ab\right)\\ =\left(a+b\right)\left(a+1\right)\left(a+b\right)\left(b+1\right)\\ =\left(a+b\right)^2\left[\left(a+1\right)\left(b+1\right)\right]\\ =\left(a+b\right)^2\left(a+b+ab+1\right)\\ =2\left(a+b\right)^2\)

Do đó phương trình đã cho tương đương:

\(\Leftrightarrow\dfrac{\left(a+b\right)}{\sqrt{2\left(a+b\right)^2}}=\dfrac{1}{\sqrt{2}}\\\Leftrightarrow\dfrac{\left(a+b\right)}{\sqrt{2}.\left(a+b\right)}=\dfrac{1}{\sqrt{2}}\left(a,b>0\right)\\ \Leftrightarrow\dfrac{1}{\sqrt{2}}=\dfrac{1}{\sqrt{2}}\left(1\right)\) 

Vì phương trình (1) đúng nên phương trình ban đầu cũng đúng

Suy ra điều phải chứng minh