Question

CMR : (a+b) \(\times\)(ab+1)  \(\ge\)4ab

Answer 1

Áp dụng bđt cosi ta có :

(a+b) . (ab+1)

>= \(2\sqrt{a.b}\).  \(2\sqrt{ab.1}\)

= \(4\sqrt{ab^2}\)

= 4ab

=> ĐPCM

Dấu "=" xảy ra <=> a = b = +-1

Tk mk nha

Answer 2

Áp dụng bất đẳng thức Cauchy ta có :

\(\left(\frac{a+b}{2}\right)^2\ge ab\ge0\)

\(\left(\frac{ab+1}{2}\right)^2\ge ab\ge0\)

\(\Leftrightarrow\left(\frac{a+b}{2}\right)^2\times\left(\frac{ab+1}{2}\right)^2\ge\left(ab\right)^2\)

\(\Leftrightarrow\left[\left(a+b\right)\left(ab+1\right)\right]^2\ge16\times\left(ab\right)^2\)

\(\Leftrightarrow\left(a+b\right)\left(ab+1\right)\ge4ab\)(đpcm)