Use of vectors

Prove trigonometry identities
Sine formula $\LARGE \frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}$
$\vec{A}+\vec{B}+\vec{C}=\vec{0}\\ \begin{align}0&=\vec{A}\times (\vec{A}+\vec{B}+\vec{C})\\ &=\vec{A}\times\vec{B}+\vec{A}\times\vec{C}\\ \vec{B}\times\vec{A}&=\vec{A}\times\vec{C}\\ ba \sin\gamma&=ac \sin\beta\\ \frac{b}{\sin\beta}&=\frac{c}{\sin\gamma}\\ \end{align}$
Similarly, expand $\vec{B}\times (\vec{A}+\vec{B}+\vec{C})$, we have $\Large \frac{a}{\sin\alpha}=\frac{c}{\sin\gamma}$ $\Box$

Cosine formula $a^2+b^2-c^2=2ab\cos\gamma$
$\begin{align}c^2&=\vec{c}\cdot\vec{c}\\ &=(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b})\\ &=\vec{a}\cdot\vec{a}+2\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{b}\\ &=a^2+2ab\cos(\pi-\gamma)+b^2\\ &=a^2-2ab\cos\gamma+b^2\:\:\Box \end{align}$

Prove inequalities

$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq \dfrac{3}{\sqrt{2}}$
Proof: Let $\vec{x}=\small\begin{pmatrix}a \\ 1-b\end{pmatrix}$, $\vec{y}=\small\begin{pmatrix}b \\ 1-c\end{pmatrix}$,$\vec{z}=\small\begin{pmatrix}c \\ 1-a\end{pmatrix}$.
By Minowski's inequality, $|\vec{x}|+|\vec{y}|+|\vec{z}| \geq |\vec{x}+\vec{y}+\vec{z}|$.
$\begin{align}\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}&\geq \sqrt{(a+b+c)^2-[3(a+b+c)]^2}\\&=\sqrt{2(a+b+c-\dfrac{3}{2})^2+\dfrac{9}{2}}\\&\geq \dfrac{3}{\sqrt{2}}\:\Box \end{align}$

$\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\geq \sqrt{6(x+y+z)}\quad x,y,z>0$
Proof: Let $\vec{a}=\small\begin{pmatrix}x \\ 1\end{pmatrix}$, $\vec{b}=\small\begin{pmatrix}y \\ 1\end{pmatrix}$,$\vec{b}=\small\begin{pmatrix}z \\ 1\end{pmatrix}$.
$\begin{align}|\vec{a}|+|\vec{b}|+|\vec{c}| &\geq |\vec{a}+\vec{b}+\vec{c}| \\ \iff \sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}&\geq \sqrt{(x+y+z)^2+3^2}\\ &\geq \sqrt{6(x+y+z)}\:\Box\end{align}$