Tìm min: $P=\frac{x+1}{y+z-1}+\frac{y+1}{z+x-1}+(\frac{x+y}{z})^2$

Question

Cho

$\left\{ \begin{array}{l} x,y,z\geq 1\\ x^2+y^2+z^2=6xy+2(x+y+z) \end{array} \right..$

Tìm min:

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

tran85295 · Answer 1 · 6/29/2016 1:35:48 PM

$gt\Leftrightarrow (x+y)^2+z^2 =8xy+2(x+y+z) \le 2(x+y)^2+2(x+y+z)$

$\Leftrightarrow (x+y)^2+2(x+y)+1 \ge z^2-2z+1$

$\Leftrightarrow (x+y+1)^2 \ge (z-1)^2\Leftrightarrow x+y+2 \ge z$

Khi đó

$VT \ge \frac{x+1}{x+2y+1}+\frac{y+1}{y+2x+1}+\frac{(x+y)^2}{(x+y+2)^2}$

$\ge \frac{(x+y+2)^2}{(x+1)(x+2y+1)+(y+1)(y+2x+1)}+\frac{(x+y)^2}{(x+y+2)^2}$

$\ge \frac{(x+y+2)^2}{(x+y)^2+4(x+y)+2xy+2}+\frac{(x+y)^2}{(x+y+2)^2} \\ \ge \frac{(x+y+2)^2}{\frac 32(x+y)^2+4(x+y)+2} +\frac{(x+y)^2}{(x+y+2)^2}$

$=\frac{2(x+y+2)}{3(x+y)+1}+\frac{(x+y)^2}{(x+y+2)^2}\overset{x+y \to t \ge 2}{=}\frac{2(t+2)}{3t+2}+\frac{t^2}{(t+2)^2} \ge \frac 54$

$P_{\min}=\frac 54\Leftrightarrow x=y=1,z=4$

Trả lời 29-06-16 01:35 PM

tran85295
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ak, qua hangouts tui hỏi :( – Confusion 29-06-16 10:00 PM

ukm, còn cách khác nhẹ nhàng hơn. – Confusion 29-06-16 09:59 PM

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