Question

Đã trả lời bởi chuyên gia

Answer 1

$Q = \frac{x + 1}{1 + y^{2}} + \frac{y + 1}{1 + z^{2}} + \frac{z + 1}{1 + x^{2}} = (\frac{x}{1 + y^{2}} + \frac{y}{1 + z^{2}} + \frac{z}{1 + x^{2}}) + (\frac{1}{1 + y^{2}} + \frac{1}{1 + z^{2}} + \frac{1}{1 + x^{2}}) = M + N$

Xét $M = \frac{x}{1 + y^{2}} + \frac{y}{1 + z^{2}} + \frac{z}{1 + x^{2}}$ , áp dụng kỹ thuật Côsi ngược dấu ta có:

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

Tương tự: $\frac{y}{1 + z^{2}} \geq y - \frac{y z}{2}$ ; $\frac{z}{1 + x^{2}} \geq z - \frac{z x}{2}$ ;

Suy ra $M = \frac{x}{1 + y^{2}} + \frac{y}{1 + z^{2}} + \frac{z}{1 + x^{2}} \geq x + y + z - \frac{x y + y z + z x}{2} = 3 - \frac{x y + y z + z x}{2}$

Lại có: $x^{2} + y^{2} + z^{2} \geq x y + y z + z x \Rightarrow {(x + y + z)}^{2} \geq 3 (x y + y z + z x) \Rightarrow x y + y z + z x \leq 3$

Suy ra: $M \geq 3 - \frac{x y + y z + z x}{2} \geq 3 - \frac{3}{2} = \frac{3}{2}$

Dấu "=" xảy ra $\Leftrightarrow x = y = z$ .

Xét $N = \frac{1}{1 + y^{2}} + \frac{1}{1 + z^{2}} + \frac{1}{1 + x^{2}}$ , ta có:

$3 - N = (1 - \frac{1}{1 + y^{2}}) + (1 - \frac{1}{1 + z^{2}}) + (1 - \frac{1}{1 + x^{2}})$

$= \frac{y^{2}}{1 + y^{2}} + \frac{z^{2}}{1 + z^{2}} + \frac{x^{2}}{1 + x^{2}} \leq \frac{y^{2}}{2 y} + \frac{z^{2}}{2 z} + \frac{x^{2}}{2 x} = \frac{x + y + z}{2} = \frac{3}{2}$

Suy ra: $N \geq 3 - \frac{3}{2} = \frac{3}{2}$ .

Dấu "=" xảy ra $\Leftrightarrow x = y = z = 1$

Từ đó suy ra: $Q \geq 3$ . Dấu "=" xảy ra $\Leftrightarrow x = y = z = 1$

Vậy $Q_{\min} = 3 \Leftrightarrow x = y = z = 1$

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Answer 2