Lý thuyết và giải bài tập về bất đẳng thức Mincopxki

\sqrt{a^2_1+b^2_1}\ +\sqrt{a^2_2+b^2_2}\ +...+\sqrt{a^2_n+b^2_n}\ge\sqrt{(a_1+a_2+...+a_n)^2+(b_1+b_2+...+b_n)^2}

\frac{a_1}{b_1}=\frac{a_2}{b_2}=...=\frac{a_n}{b_n}

\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge \sqrt{(a+c)^2+(b+d)^2}

\frac{a}{b}=\frac{c}{d}

\sqrt{a^2+b^2}+\sqrt{c^2+d^2}+\sqrt{e^2+f^2}\ge \sqrt{(a+c+e)^2+(b+d+f)^2}

\frac{a}{b}=\frac{c}{d}=\frac{e}{f}

\sqrt{a^2+x^2}+\sqrt{b^2+y^2}\ge \sqrt{(a+b)^2+(x+y)^2}

\begin{aligned}
&\sqrt{a^2+x^2}+\sqrt{b^2+y^2}\ge \sqrt{(a+b)^2+(x+y)^2}\\
&\Leftrightarrow a^2+x^2+b^2+y^2+2\sqrt(a^2+x^2)(b^2+y^2)\ge a^2+x^2+b^2+y^2+2ab+2xy\\
&\Leftrightarrow 2\sqrt{(a^2+x^2)(b^2+y^2)}\ge 2ab+2xy\\
&\Leftrightarrow \sqrt{(a^2+x^2)(b^2+y^2)}\ge ab+xy \ (*)
\end{aligned}

\begin{aligned}
&\footnotesize \text{Đặt }\vec{u}=(a;x) \text{ và }\vec{v}=(b;y). \text{ Khi đó }\vec{u}+\vec{v}=(a+b;x+y).\\
&\footnotesize \text{Từ bất đẳng thức véc tơ }|\vec{u}+\vec{v}|\le |\vec{u}|+|\vec{v}| \text{ và công thức độ dài vectơ ta có ngay điều phải chứng minh.}
\end{aligned}

\sqrt{a^2+x^2}+\sqrt{b^2+y^2}+\sqrt{c^2+z^2}\ge\sqrt{(a+b+c)^2+(x+y+z)^2}

\frac{\sqrt{b^2+2a^2}}{ab}+\frac{\sqrt{c^2+2b^2}}{bc}+\frac{\sqrt{a^2+2c^2}}{ca}\ge\sqrt3

ab+bc+ca=abc\Leftrightarrow\frac1a+\frac1b+\frac1c=1

\frac{\sqrt{b^2+2a^2}}{ab}+\frac{\sqrt{c^2+2b^2}}{bc}+\frac{\sqrt{a^2+2c^2}}{ca}=\sqrt{\frac{1}{a^2} + \frac{2}{b^2}}+\sqrt{\frac{1}{b^2} + \frac{2}{c^2}}+\sqrt{\frac{1}{c^2} + \frac{2}{a^2}}

\begin{aligned}
&\small\sqrt{\frac{1}{a^2}+\left(\frac{\sqrt2}{b}\right)^2}+\sqrt{\frac{1}{b^2}+\left(\frac{\sqrt2}{c}\right)^2}+\sqrt{\frac{1}{c^2}+\left(\frac{\sqrt2}{a}\right)^2} \ge \sqrt{\left(\frac1a+\frac1b+\frac1c\right)^2 +2\left(\frac1a+\frac1b+\frac1c\right)^2}\\
&\text{Mà: }\frac1a+\frac1b+\frac1c =1 \Rightarrow \frac{\sqrt{b^2+2a^2}}{ab}+\frac{\sqrt{c^2+2b^2}}{bc}+\frac{\sqrt{a^2+2c^2}}{ac} \ge\sqrt3
\end{aligned}

z=a+bi\ (a,b\in\R) \text{ thỏa mãn } |z-4-3i|=|\overline{z}-2+1|

P = a^2 + b^2 \text{ khi }| z + 1 - 3i | + | z - 1 + i | \text{ đạt giá trị nhỏ nhất.}

\small \begin{aligned}
(a-4)^2 + (b-3)^2 &= (a-2)^2 + (1-b)^2 ⇔ b = 5-a\\
|z+1-3i|+|z-1+i|&=\sqrt{(a+1)^2+(b-3)^2}+\sqrt{(a-1)^2+(b+1)^2}\\
&=\sqrt{(a+1)^2+(2-a)^2}+\sqrt{(a-1)^2+(6-a)^2}\\
&=\sqrt{2a^2-2a+5}+\sqrt{2a^2-14a+37}\\
&=\sqrt{\left(\frac{1}{\sqrt2}-\sqrt2a \right)^2+\left(\sqrt{\frac92}\right)^2}+\sqrt{\left(\sqrt2a-\frac{7}{\sqrt2}\right)^2+\left(\sqrt{\frac{25}{2}}\right)^2}\\
&=\sqrt{\left(\frac{1}{\sqrt2}-\sqrt2a+\sqrt2a-\frac{7}{\sqrt2} \right)^2+ \left(\sqrt{\frac92}+\sqrt{\frac{25}{2}} \right)^2}=5\sqrt2
\end{aligned}

\frac{\frac{1}{\sqrt2}-\sqrt2a}{\sqrt2a-\frac{7}{\sqrt2}}=\frac{\sqrt{\frac92}}{\sqrt{\frac{25}{2}}}\Leftrightarrow 
\begin{cases}a=\frac{13}{8}\\b=\frac{27}{8}\end{cases} \Rightarrow P=\frac{13^2+27^2}{8^2}=\frac{449}{32}

\small \begin{aligned}
MA+2MB&=\sqrt{(x-7)^2+(y-9)^2+z^2}+2\sqrt{x^2+(y-8)^2+z^2}\\
&=\sqrt{(x-7)^2+(y-9)^2+z^3+3[(x-1)^2+(y-1)^2+z^2-25]}+2\sqrt{x^3+(y-8)^2+z^2}\\
&=2\left[\sqrt{\left(\frac52-x \right)^2+(3-y)^2+(-z)^2}+\sqrt{x^2+(y-8)^2+z^2}\right]\\
&\ge 2\sqrt{\left( \frac52-x+x\right)^2+(3-y+8-y)^2+(-z+z)^2}=5\sqrt5
\end{aligned}

\begin{cases} \frac{\frac52-x}{x}=\frac{3-y}{y-8}=k >0\\ z=0\\ (x-1)^2+(y-1)^2+z^2=25 \end{cases} \Leftrightarrow\begin{cases} x=1\\ y=6\\z=0\end{cases}\Leftrightarrow M(1;6;0)