Tim Min P=$ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{\sqrt{a^2+b^2+c^2}}$

Question

Cho $a,b,c >0$ tm $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{2c^{2}}$.

Lmdi mn. Mk pai kien tri lam ms nhap dc xong cau hoi nay. Jup vs nhe

Nguyễn Nhung · Answer 1 · 7/1/2016 11:19:12 AM

Ta có: $\frac{c^{2}}{a^{2}} + \frac{c^{2}}{b^{2}} = \frac{1}{2}; \frac{a}{c} = x; \frac{b}{c} = y \Rightarrow \frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{1}{2}$ $\Rightarrow 1 \geq (\frac{1}{x} + \frac{1}{y})^{2} \Leftrightarrow x y \geq x + y$

$P = \frac{x}{y + 1} + \frac{y}{x + 1} + \frac{1}{\sqrt{x^{2} + y^{2} + 1}}$

Chứng minh: $x^{2} + y^{2} + 1 \leq (x + y - 1)^{2} \Leftrightarrow 2 x y - 2 x - 2 y \geq 0 \Leftrightarrow x y \geq x + y (d u n g)$

suy ra: $P \geq \frac{x}{y + 1} + \frac{y}{x + 1} + \frac{1}{x + y - 1} \geq \frac{(x + y)^{2}}{2 x y + x + y} + \frac{1}{x + y - 1} \geq \frac{(x + y)^{2}}{x + y + \frac{(x + y)^{2}}{2}} + \frac{1}{x + y - 1} = \frac{2 (x + y)}{x + y + 2} + \frac{1}{x + y - 1} = \frac{2 a}{a + 2} + \frac{1}{a - 1} = A$

Lại có; $A = \frac{2 a}{a + 2} + \frac{1}{a - 1} = \frac{2 a^{2} - 2 a + a + 2}{(a - 1) (a + 2)} = \frac{2 a^{2} - a + 2}{a^{2} + a - 2} \Leftrightarrow a^{2} (A - 2) + a (A + 1) - 2 A - 2 = 0; Δ = A^{2} + 2 A + 1 + 4 (A - 2) (2 A + 2) \geq 0 \Leftrightarrow A \geq \frac{5}{3}$

Dấu = xảy ra khi ${\begin{matrix} \frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{1}{2} \\ \frac{x}{y + 1} = \frac{y}{x + 1} \\ x + y = x y \\ a = x + y = 4 \end{matrix} \Leftrightarrow {\begin{matrix} c = 2 a \\ c = 2 b \end{matrix}$