Cực trị

L

lp_qt

$\dfrac{1}{2}x^2+\dfrac{17-\sqrt{33}}{16}y^2 \ge 2.\sqrt{\dfrac{17-\sqrt{33}}{32}}xy$

$\dfrac{1}{2}x^2+\dfrac{17-\sqrt{33}}{16}z^2 \ge 2.\sqrt{\dfrac{17-\sqrt{33}}{32}}xz$

$\dfrac{-1+\sqrt{33}}{16}y^2+\dfrac{-1+\sqrt{33}}{16}z^2 \ge 2. \dfrac{-1+\sqrt{33}}{16}yz$

\Rightarrow $x^2+y^2+z^2 \ge 4.\dfrac{-1+\sqrt{33}}{16}(xy+xz+2zy)$

\Rightarrow $P \ge \dfrac{-1+\sqrt{33}}{4}$
 
H

huynhbachkhoa23

Đặt $k=\dfrac{\sqrt{3}+1}{2}$ thì $k=\dfrac{2k+1}{2k}$
Ta luôn có $(y-z)^2\ge 0$ nên $y+z\le \sqrt{2(y^2+z^2)}$ và $2yz\le y^2+z^2$
Do đó $x(y+z)+2yz\le \sqrt{2x^2(y^2+z^2)}+y^2+z^2=\sqrt{2kx^2.\dfrac{y^2+z^2}{k}}+y^2+z^2\le \dfrac{2kx^2+\dfrac{y^2+z^2}{k}}{2}+y^2+z^2=kx^2+\dfrac{2k+1}{2k}(y^2+z^2)=k(x^2+y^2+z^2)$
Vậy ta có $\dfrac{x^2+y^2+z^2}{xy+2yz+zx}\ge \dfrac{x^2+y^2+z^2}{k(x^2+y^2+z^2)}=\dfrac{1}{k}= \sqrt{3}-1$
 
You must log in or register to reply here.
Top Bottom