Bài 5: Ứng dụng chứng minh bất đẳng thức, tìm cực trị

I. Lý thuyết 
$\vec{a}=(x_1,y_1,z_1)$
$\vec{b}=(x_2,y_2,z_2)$
1) $\left | \vec{a},\vec{b} \right |\leq \left | \vec{a} \right |.\left | \vec{b} \right |$
$\Rightarrow \left | x_1x_2+y_1y_2+z_1z_2 \right |\leq \sqrt{x^2_1+y^2_1+z_1^2}.\sqrt{x^2_2+y^2_2+z_2^2}$
2) $\left | \vec{a}+\vec{b} \right |\leq \left | \vec{a} \right |+\left | \vec{b} \right |$
$\Rightarrow \sqrt{(x_1+x_2)^2+(y_1+y_2)^2+(z_1+z_2)^2}\leq \sqrt{x^2_1+y^2_1+z^2_1}+\sqrt{x^2_2+y^2_2+z^2_2}$
3) $\left | \vec{a} +\vec{b} +\vec{c} \right |\leq \left | \vec{a} \right |+ \left | \vec{b} \right |+\left | \vec{c} \right |$
$\Rightarrow \sqrt{(x_1+x_2+x_3)^2+(y_1+y_2+y_3)^2+(z_1+z_2+z_3)^2}$
$\leq \sqrt{x^2_1+y_1^2+z_1^2}+\sqrt{x^2_2+y_2^2+z_2^2}+ \sqrt{x^2_3+y_3^2+z_3^2}$
Dấu đẳng thức xảy ra khi $\vec{a},\vec{b},\vec{c}$ cùng hướng
II. Bài tập
VD1: Tìm GTNN của biểu thức
$T=\sqrt{x^2+y^2+z^2}+\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}$ với $x,y,z\neq 0$
Giải
Xét $\vec{a}=(x;y;z)\Rightarrow \left | \vec{a} \right |=\sqrt{x^2+y^2+z^2}$
$\vec{b}=\left ( \frac{1}{x};\frac{1}{y};\frac{1}{z} \right ) \Rightarrow \left | \vec{b} \right |=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}$
$\vec{a}+\vec{b}=\left ( x+\frac{1}{x}; y+\frac{1}{y} ;z+\frac{1}{z} \right )$
$\left | \vec{a}+\vec{b} \right |=\sqrt{(x+\frac{1}{x})^2+(y+\frac{1}{y})^2 +(z+\frac{1}{z})^2}$
Do $(x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2\geq 2+2=4$
Tương tự $(y+\frac{1}{y})^2\geq 4$
$(z+\frac{1}{z})^2\geq 4$
$\Rightarrow \left | \vec{a}+\vec{b} \right |\geq \sqrt{12}=2\sqrt{3}$
Ta có $\Rightarrow \left | \vec{a} \right |+\left | \vec{b} \right |\geq \left | \vec{a} +\vec{b}\right |$
$\Rightarrow \sqrt{x^2+y^2+z^2}+\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}} \geq 2\sqrt{3}$
$T\geq 2\sqrt{3}$
GTNN $T\geq 2\sqrt{3}$ 
Dấu đẳng thức xảy ra khi
$\left\{\begin{matrix} \vec{a}, \vec{b} \ cung \ huong\\ x^2=y^2=z^2=1 \end{matrix}\right.\Rightarrow \bigg \lbrack\begin{matrix} x=y=z=1\\ x=y=z=-1 \end{matrix}$
VD2: Cho $x+2y+2z=9$. Tìm GTNN của $T=x^2+y^2+z^2$
Giải
Xét $\vec{a}=(1;2;2)\Rightarrow \left | \vec{a} \right |=\sqrt{1^2+2^2+2^2}=3$
$\vec{b}=(x;y;z)\Rightarrow \left | \vec{b} \right |=\sqrt{z^2+y^2+z^2}$
$\vec{a}.\vec{b}=x+2y+2z=9$
Ta có $\vec{a}.\vec{b}\leq \left | \vec{a} \right |.\left | \vec{b} \right |$
$\Rightarrow 9\leqslant 3.\sqrt{x^2+y^2+z^2}$
$\Rightarrow 9\leqslant x^2+y^2+z^2=T$
GTNN = 9 khi $\vec{a}, \vec{b}$ cùng hướng mà $\vec{a}\neq \vec{0},\exists k\geq 0, \vec{b}=k.\vec{a}$
$\left\{\begin{matrix} x=k\\ y=2k\\ z=2k\\ x+2y+2z=9 \end{matrix}\right.\Rightarrow \left\{\begin{matrix} k=1\\ x=1\\ y=2\\ z=2 \end{matrix}\right.$
Vậy GTNN T = 9 khi x = 1, y = 2, z = 2
VD3: Cho $(x-1)^2+(y-2)^2+(z-3)^2=9$. Tìm GTNN, GTLN của T = x + 2y + 2z
Giải
Xét 
$\vec{a}=(x-1;y-2;z-3)\Rightarrow \left | \vec{a} \right |=\sqrt{(x-1)^2+(y-2)^2+(z-3)^2}=3$
$\vec{b}=(1;2;2)\Rightarrow \left | \vec{b} \right |=\sqrt{1^2+2^2+2^2}=3$
$\vec{a}.\vec{b}=1(x-1)+2(y-2)+2(z-3)$
$=x+2y+2z-11$
Ta có $\left | \vec{a}.\vec{b} \right |\leq \left | \vec{a} \right |.\left | \vec{b} \right |$
$\Leftrightarrow \left | x+2y+2z-11 \right |\leq 9$
$\Leftrightarrow -9\leq x+2y+2z-11\leq 9$
$\Leftrightarrow 2\leq x+2y+2z\leq 20$
$\Leftrightarrow 2\leq T \leq 20$
GTLN T = 20 khi $\left\{\begin{matrix} x-1=\frac{y-2}{2}=\frac{z-3}{2}=\frac{z-3}{2}=k\geq 0\\ (x-1)^2+(y-2)^2+(z-3)^2=9 \end{matrix}\right.$
$\Rightarrow \left\{\begin{matrix} x-1=k\geq 0\\ y-2=2k\\ z-3=2k\\ 9k^2=9 \end{matrix}\right.\Rightarrow \left\{\begin{matrix} k=1\\ x=2\\ y=4\\ z=5 \end{matrix}\right.$
GTNN T = 2 khi x = 0, y = 0, z = 1
VD4: Cho $x+y+z\leq 1, x,y,z>0$. Tìm GTNN của $T=\sqrt{x^2+\frac{1}{x^2}+1}+\sqrt{y^2+\frac{1}{y^2}+1}+\sqrt{z^2+\frac{1}{z^2}+1}$
Giải
Xét
$\vec{a}=(x;\frac{1}{x};1)\Rightarrow \left | \vec{a} \right |= \sqrt{x^2+\frac{1}{x^2}+1}$
$\vec{b}=(y;\frac{1}{y};1)\Rightarrow \left | \vec{b} \right |= \sqrt{y^2+\frac{1}{y^2}+1}$
$\vec{c}=(z;\frac{1}{z};1)\Rightarrow \left | \vec{c} \right |= \sqrt{z^2+\frac{1}{z^2}+1}$
$\vec{a}+\vec{b}+\vec{c}=(x+y+z;\frac{1}{x}+\frac{1}{y};\frac{1}{z};3)$
$\left | \vec{a}+\vec{b}+\vec{c} \right |= \sqrt{(x+y+z)^2+(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2+9}$
$\geq \sqrt{(x+y+z)^2+\frac{81}{(x+y+z)^2}+9}=T_1$
Đặt $t=(x+y+z)^2, 0<t\leq 1$
$T_1=\sqrt{t+\frac{81}{t}+9}=\sqrt{(t+\frac{1}{t})+\frac{80}{t}+9}$
$\geq \sqrt{2+80+9}=\sqrt{91}$
Ta có $\left | \vec{a} \right |+\left | \vec{b} \right |+\left | \vec{c} \right | \geq \left | \vec{a} + \vec{b} +\vec{c}\right |$
$T\geq \sqrt{91}$
GTNN $T\geq \sqrt{91}$, đạt được khi $x=y=z=\frac{1}{3}$