a.
Biến đổi tương đương:
\(\dfrac{a^2+b^2}{2}\ge ab+\dfrac{\left(a-b\right)^2}{a^2+b^2+2}\)
\(\Leftrightarrow\dfrac{a^2+b^2}{2}-ab-\dfrac{\left(a-b\right)^2}{a^2+b^2+2}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{2}-\dfrac{\left(a-b\right)^2}{a^2+b^2+2}\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(\dfrac{1}{2}-\dfrac{1}{a^2+b^2+2}\right)\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2\left(a^2+b^2\right)}{2\left(a^2+b^2+2\right)}\ge0\) (luôn đúng)
b.
\(Q=-6\left(a+b\right)+5\left(a+\dfrac{4}{a}\right)+7\left(b+\dfrac{1}{b}\right)\)
\(Q\ge-6.3+5.2\sqrt{\dfrac{4a}{a}}+7.2\sqrt{\dfrac{b}{b}}=16\)
\(Q_{min}=16\) khi \(\left(a;b\right)=\left(2;1\right)\)