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

Xét \(k = 2 m\) thì \(n = 6 m\) suy ra \(n \left(\right. n + 1 \left.\right) = 6 m \left(\right. 6 m + 1 \left.\right)\) chia hết cho \(6\).

 Xét \(k = 2 m + 1\) thì \(n = 3 \left(\right. 2 m + 1 \left.\right) = 6 m + 3\).

Suy ra \(n \left(\right. n + 1 \left.\right) = \left(\right. 6 m + 3 \left.\right) \left(\right. 6 m + 4 \left.\right) = 3. \left(\right. 2 m + 1 \left.\right) . 2 \left(\right. 3 m + 2 \left.\right) = 6. \left(\right. 2 m + 1 \left.\right) . \left(\right. 3 m + 2 \left.\right)\) chia hết cho \(6\).

Vậy với mọi số tự nhiên \(n\), nếu \(n\) chia hết cho \(3\) thì \(n \left(\right. n + 1 \left.\right)\) chia hết cho \(6\).

Nếu \(n\) lẻ thì \(n\) có dạng \(n = 2 k + 1\) với \(k \in \mathbb{N}\).

Do đó \(n^{3} = \left(\right. 2 k + 1 \left.\right)^{3} = 8 k^{3} + 12 k^{2} + 6 k + 1 = 2 \left(\right. 4 k^{3} + 6 k^{2} + 3 k \left.\right) + 1\).

Suy ra \(n^{3}\) lẻ.