Question

Đã trả lời bởi chuyên gia

Answer 1

Ta có:

$\begin{array}{l} \frac{1}{a^{2}} + \frac{1}{b^{2}} \geq 2 \cdot \frac{1}{a} \cdot \frac{1}{b} \\ \Leftrightarrow 2 (\frac{1}{a^{2}} + \frac{1}{b^{2}}) \geq \frac{1}{a^{2}} + \frac{1}{b^{2}} + 2 \cdot \frac{1}{a} \cdot \frac{1}{b} \\ \Leftrightarrow 2 (\frac{1}{a^{2}} + \frac{1}{b^{2}}) \geq {(\frac{1}{a} + \frac{1}{b})}^{2} (1) \end{array}$

Mặt khác:

$\begin{array}{l} {(\frac{1}{a} + \frac{1}{b} - 2)}^{2} \geq 0 \\ \Leftrightarrow {(\frac{1}{a} + \frac{1}{b})}^{2} - 4 (\frac{1}{a} + \frac{1}{b}) + 4 \geq 0 \\ \Leftrightarrow {(\frac{1}{a} + \frac{1}{b})}^{2} + 4 \geq 4 (\frac{1}{a} + \frac{1}{b}) (2) \end{array}$

Lại có: (a + b)² ≥ 4ab

\Leftrightarrow \frac{a + b}{a b} \geq \frac{4}{a + b} \Leftrightarrow \frac{1}{a} + \frac{1}{b} \geq \frac{4}{a + b} (3)

Từ (1), (2), (3) ⇒ $2 (\frac{1}{a^{2}} + \frac{1}{b^{2}}) + 4 \geq {(\frac{1}{a} + \frac{1}{b})}^{2} + 4 \geq 4 (\frac{1}{a} + \frac{1}{b}) \geq \frac{16}{a + b}$

$\Leftrightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} + 2 \geq \frac{8}{a + b}$ (với a, b > 0)

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Answer 2