TITLE: An inequality related to the cosine theorem
QUESTION [6 upvotes]: Let $A,B,C$ be the three angles in the a triangle (with length $a,b,c$). Can we show that 
$$x^2+y^2+z^2\geq 2x y \cos A+2xz\cos B+2yz\cos C?$$
for all $x,y,z\in\Bbb R$.
I do not see whether it is true, but I find it in some book.
I suspect it is false. My idea is that if $x=a$, $y=b$, $z=c$, then the cosine theorem implies that the inequality is in fact equality...
Hence the matrix 
$$\begin{pmatrix}1&-\cos A&-\cos B\\-\cos A&1&-\cos C\\ -\cos B&-\cos C&1\end{pmatrix}$$
is positive semidefinite. 
However, there may be some minors (say $-\cos A$) be negative, contradicting to the fact that $M$ is positive semi-definite $\Leftrightarrow$ all minors $\geq 0$...

REPLY [2 votes]: Let our triangle is $\Delta ABC$. Let $\vec i=\frac{1}{AB}\vec{AB}$, $\vec{j}=\frac{1}{CA}\vec{CA}$ and $\vec{k}=\frac{1}{BC}\vec{BC}$.
Hence, our inequality is $\left(x\vec{i}+y\vec{j}+z\vec{k}\right)^2\geq0$.
The last inequality it's just
$$\sum_{cyc}(x^2+2xy\vec{i}\vec{j})\geq0$$ or
$$\sum_{cyc}(x^2+2xy\cos(180^{\circ}-A))\geq0$$ or
$$x^2+y^2+z^2\geq2xy\cos{A}+2yz\cos{B}+2zx\cos{C}.$$