Ta có \(\frac{a x + b y}{2} \geq \&\text{nbsp}; \frac{a + b}{2} . \&\text{nbsp}; \frac{x + \&\text{nbsp}; y}{2}\)

\(\Leftrightarrow 2 \left(\right. a x + b y \left.\right) \&\text{nbsp}; \geq \&\text{nbsp}; \left(\right. a + b \left.\right) \left(\right. x + y \left.\right)\)

\(\Leftrightarrow 2 \left(\right. a x + b y \left.\right) \&\text{nbsp}; \geq a x + a y + b x + b y\)

\(\Leftrightarrow a x + b y - a y - b x \&\text{nbsp}; \geq 0\)

\(\Leftrightarrow \left(\right. a - b \left.\right) \left(\right. x - y \left.\right) \geq 0\) (luôn đúng vì giả thiết \(a \geq b\) và \(x \geq y\)).

Vậy nếu \(a \geq b\), \(x \geq y\) thì \(\frac{a x + b y}{2} \geq \&\text{nbsp}; \frac{a + b}{2} . \&\text{nbsp}; \frac{x + \&\text{nbsp}; y}{2}\).

Ta có \(4 x^{2} + 4 y^{2} + 6 x + 3 \geq 4 x y\)

\(\Leftrightarrow \left(\right. x^{2} - 4 x y + 4 y^{2} \left.\right) + 3 \left(\right. x^{2} + 2 x + 1 \left.\right) \geq 0\)

\(\Leftrightarrow \&\text{nbsp}; \left(\right. x - 2 y \left.\right)^{2} + 3 \left(\right. x \&\text{nbsp}; + 1 \left.\right)^{2} \geq 0\) (luôn đúng với mọi \(x\), \(y\)).

Vậy với mọi \(x\), \(y\) ta có \(4 x^{2} + 4 y^{2} + 6 x + 3 \geq 4 x y\).

Ta có \(4 x^{2} + 4 y^{2} + 6 x + 3 \geq 4 x y\)

\(\Leftrightarrow \left(\right. x^{2} - 4 x y + 4 y^{2} \left.\right) + 3 \left(\right. x^{2} + 2 x + 1 \left.\right) \geq 0\)

\(\Leftrightarrow \&\text{nbsp}; \left(\right. x - 2 y \left.\right)^{2} + 3 \left(\right. x \&\text{nbsp}; + 1 \left.\right)^{2} \geq 0\) (luôn đúng với mọi \(x\), \(y\)).

Vậy với mọi \(x\), \(y\) ta có \(4 x^{2} + 4 y^{2} + 6 x + 3 \geq 4 x y\).

Ta có \(4 x^{2} + 4 y^{2} + 6 x + 3 \geq 4 x y\)

\(\Leftrightarrow \left(\right. x^{2} - 4 x y + 4 y^{2} \left.\right) + 3 \left(\right. x^{2} + 2 x + 1 \left.\right) \geq 0\)

\(\Leftrightarrow \&\text{nbsp}; \left(\right. x - 2 y \left.\right)^{2} + 3 \left(\right. x \&\text{nbsp}; + 1 \left.\right)^{2} \geq 0\) (luôn đúng với mọi \(x\), \(y\)).

Vậy với mọi \(x\), \(y\) ta có \(4 x^{2} + 4 y^{2} + 6 x + 3 \geq 4 x y\).