Question

[Mai Thành Đạt]

Cho hai số a,b thỏa mãn \(ab\ge1\)

Chứng minh rằng:\(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\)

Answer 1

[Mỹ Duyên]

Dùng phương pháp biến đổi tương đương nhé!!!

Ta có : \(\dfrac{1}{1+a^2}\) + \(\dfrac{1}{1+b^2}\) \(\ge\) \(\dfrac{2}{1+ab}\)

<=>( \(\dfrac{1}{1+a^2}\) - \(\dfrac{1}{1+ab}\) ) + ( \(\dfrac{1}{1+b^2}\) - \(\dfrac{1}{1+ab}\) ) \(\ge\) 0

<=> \(\dfrac{1+ab-1-a^2}{\left(1+a^2\right)\left(1+ab\right)}\) + \(\dfrac{1+ab-1-b^2}{\left(1+b^2\right)\left(1+ab\right)}\) \(\ge\) 0

<=> \(\dfrac{ab-a^2}{\left(1+a^2\right)\left(1+ab\right)}\) + \(\dfrac{ab-b^2}{\left(1+b^2\right)\left(1+ab\right)}\) \(\ge\) 0

<=> \(\dfrac{a\left(b-a\right)\left(1+b^2\right)+b\left(a-b\right)\left(1+a^2\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\) \(\ge\) 0

<=> \(a\left(b-a\right)\left(1+b^2\right)-b\left(b-a\right)\left(1+a^2\right)\) \(\ge\) 0

<=> \(\left(b-a\right)\left(a+ab^2-b-a^2b\right)\) \(\ge\) 0

<=> \(\left(b-a\right)\left[ab\left(b-a\right)-\left(b-a\right)\right]\) \(\ge\) 0

<=> \(\left(b-a\right)\left(b-a\right)\left(ab-1\right)\) \(\ge\) 0

<=> \(\left(b-a\right)^2\left(ab-1\right)\) \(\ge\) 0 (1)

Mà \(\left\{{}\begin{matrix}\left(b-a\right)^2\ge0\\ab-1\ge0\end{matrix}\right.\) ( vì ab \(\ge\)1)

=> \(\left(b-a\right)^2\left(ab-1\right)\) \(\ge\) 0

=> (1) luôn đúng

Vậy đpcm ....

Answer 2

[BW_P&A]

Ta có: \(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\)

\(\Leftrightarrow\left(\dfrac{1}{1+a^2}-\dfrac{1}{1+b^2}\right)+\left(\dfrac{1}{1+b^2}-\dfrac{1}{1+ab}\right)\ge0\)

\(\Leftrightarrow\dfrac{ab-a^2}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{ab-b^2}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Leftrightarrow\dfrac{a\left(b-a\right)}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{b\left(a-b\right)}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(b-a\right)^2\left(ab-1\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\)

BĐT cuối cùng đúng vì \(a.b\ge1\Rightarrowđpcm\)