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 \(x=y=z=\frac{1}{3}\)