Question

Đã trả lời bởi chuyên gia

Answer 1

Ta có: (x – y)² + (y – z)² + (z – x)² ≥ 0

⇔ x² – 2xy + y² + y² – 2yz + z² + z² – 2xz + x² ≥ 0

⇔ 2x² – 2xy + 2y² – 2yz + 2z² – 2xz ≥ 0

⇔ x² + y² + z² – xy – yz – xz ≥ 0

⇔ x² + y² + z² + 2xy + 2yz + 2xz ≥ 3xy + 3yz + 3xz

⇔ (x + y + z)² ≥ 3(xy + yz + xz)

\Leftrightarrow x y + y z + x z \leq \frac{{(x + y + z)}^{2}}{3} = \frac{3^{2}}{3} = 3

Ta có

\frac{x + 1}{y^{2} + 1} = (x + 1) . \frac{1}{y^{2} + 1} = (x + 1) . (1 - \frac{y^{2}}{y^{2} + 1}) \geq (x + 1) (1 - \frac{y^{2}}{2 y}) = x + 1 - \frac{y (x + 1)}{2}

\frac{y + 1}{z^{2} + 1} = (y + 1) . \frac{1}{z^{2} + 1} = (y + 1) . (1 - \frac{z^{2}}{z^{2} + 1}) \geq (y + 1) . (1 - \frac{z^{2}}{2 z}) = y + 1 - \frac{z (y + 1)}{2}

\frac{z + 1}{x^{2} + 1} = (z + 1) . \frac{1}{x^{2} + 1} = (z + 1) . (1 - \frac{x^{2}}{x^{2} + 1}) \geq (z + 1) . (1 - \frac{x^{2}}{2 x}) = z + 1 - \frac{x (z + 1)}{2}

Suy ra $P \geq x + 1 - \frac{y (x + 1)}{2} + y + 1 - \frac{z (y + 1)}{2} + z + 1 - \frac{x (z + 1)}{2}$

P \geq x + y + z + 3 - \frac{y (x + 1) + z (y + 1) + x (z + 1)}{2}

P \geq 6 - \frac{(x + y + z) + (x y + y z + x z)}{2}

P \geq 6 - \frac{3 + 3}{2} = 3

Dấu “ = ” xảy ra khi x² = y² = z² = 1 hay x = y = 1

Vậy P đạt giá trị nhỏ nhất bằng 3 khi x = y = z = 1.

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Answer 2