Question

[pham trung thanh]

Cho a;b là các số thực thỏa mãn điều kiện \(a+b\ne0\). Chứng minh rằng

\(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge2\)

Answer 1

[ɱ√ρ︵ʙᴇsт➻❥ɴᴀκᴀʀoтн★彡]

Ta có: \(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge2\)

\(\Leftrightarrow\left(a^2+b^2\right)\left(a+b\right)^2+\left(ab+1\right)^2\ge2\left(a+b\right)^2\)

\(\Leftrightarrow\left(a+b\right)^2\left[\left(a+b\right)^2-2ab\right]-2\left(a+b\right)^2+\left(ab+1\right)^2\ge0\)

\(\Leftrightarrow\left(a+b\right)^4-2ab\left(a+b\right)^2-2\left(a+b\right)^2+\left(ab+1\right)^2\ge0\)

\(\Leftrightarrow\left[\left(a+b\right)^2-ab-1\right]^2\ge0\)(đúng) 

\(\Leftrightarrow dpcm\)

Answer 2

[phạm văn tuấn]

⇔(a2+b2)(a+b)2+(ab+1)2≥2(a+b)2

⇔(a+b)2[(a+b)2−2ab]−2(a+b)2+(ab+1)2≥0

⇔(a+b)4−2ab(a+b)2−2(a+b)2+(ab+1)2≥0

⇔[(a+b)2−ab−1]2≥0(đúng) 

           k mình đi