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