Ta có :\(\left(a+b\right)^2+\dfrac{a+b}{2}=\left(a+b\right)\left(a+b+\dfrac{1}{2}\right)\)
=\(\left(a+b\right)\left(a+\dfrac{1}{4}+b+\dfrac{1}{4}\right)\)
Áp dụng bđt cô si ta có:
a+b\(\ge2\sqrt{ab}\),\(a+\dfrac{1}{4}\ge\sqrt{a},b+\dfrac{1}{4}\ge\sqrt{b}\)
do đó \(\left(a+b\right)^2+\dfrac{\left(a+b\right)}{2}\ge2\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)=2a\sqrt{b}+2b\sqrt{a}\)
Dấu "=" xảy ra khi:a=b=\(\dfrac{1}{4}\)
Vậy với a,b là các số thực dương ta có \(\left(a+b\right)^2+\dfrac{a+b}{2}\ge2a\sqrt{b}+2b\sqrt{a}\)
\(\left(a+b\right)^2+\dfrac{a+b}{2}=\left(a+b\right)\left(a+b+\dfrac{1}{2}\right)\)
\(=\left(a+b\right)\left[\left(a+\dfrac{1}{4}\right)\left(b+\dfrac{1}{4}\right)\right]\ge2\sqrt{ab}\left(a+b\right)=2a\sqrt{b}+2b\sqrt{a}\)
