Question

[Trần Thị Thu Ngân]

Chứng minh BĐT :

\(\dfrac{\left(a+b\right)^2}{2}+\dfrac{a+b}{4}\ge a\sqrt{b}+b\sqrt{a}\) với a,b\(\ge\)0

Answer 1

[Trần Thiên Kim]

Áp dụng bđt Cô-si chi 2 số không âm, ta có:\(\dfrac{\left(a+b\right)^2}{2}+\dfrac{a+b}{4}=\dfrac{a+b}{2}\left(a+b+\dfrac{1}{2}\right)\ge\sqrt{ab}\left(a+b+\dfrac{1}{2}\right)\)

Xét \(\sqrt{ab}\left(a+b+\dfrac{1}{2}\right)\ge a\sqrt{b}+b\sqrt{a}\)

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

\(\Leftrightarrow a+b+\dfrac{1}{2}\ge\sqrt{a}+\sqrt{b}\)

\(\Leftrightarrow a-\sqrt{a}+\dfrac{1}{4}+b-\sqrt{b}+\dfrac{1}{4}\ge0\)

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

\(\Rightarrow\sqrt{ab}\left(a+b+\dfrac{1}{2}\right)\ge a\sqrt{b}+b\sqrt{a}\)

Mà \(\dfrac{\left(a+b\right)^2}{2}+\dfrac{a+b}{4}\ge\sqrt{ab}\left(a+b+\dfrac{1}{2}\right)\)

\(\Rightarrow\dfrac{\left(a+b\right)^2}{2}+\dfrac{a+b}{4}\ge a\sqrt{b}+b\sqrt{a}\)

Answer 2

[Trần Thị Thu Ngân]

cảm ơn bạn nhiều !