An inequality related to the cosine theorem

Question

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 $\geq 0$...

Answer 1

Notice that

$x^2+y^2+z^2- 2x y \cos A-2xz\cos B-2yz\cos C=$

$$(x,y,z)\begin{pmatrix}1&-\cos A&-\cos B\\-\cos A&1&-\cos C\\ -\cos B&-\cos C&1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}.$$

Now, since $A+B+C=\pi$ then $(\pi-A)+(\pi-B)+(\pi-C)=2\pi$.

Thus, we can consider three normalized vectors $v_1,v_2,v_3$ in $\mathbb{R}^2$ with origin in $(0,0)$ such that the angle between $v_1$ and $v_2$ is $\pi-A$, the angle between $v_2$ and $v_3$ is $\pi-C$ and the angle between $v_3$ and $v_1$ is $\pi-B$.

Now, conside the Gram matrix $G(v_1,v_2,v_3)=\begin{pmatrix}\langle v_1,v_1 \rangle& \langle v_1,v_2 \rangle& \langle v_1,v_3 \rangle\\\langle v_2,v_1 \rangle& \langle v_2,v_2 \rangle& \langle v_2,v_3 \rangle\\ \langle v_3,v_1 \rangle& \langle v_3,v_2 \rangle& \langle v_3,v_3 \rangle\end{pmatrix}$.

Notice that $\langle v_1,v_1 \rangle=\langle v_2,v_2 \rangle=\langle v_3,v_3 \rangle=1$.

Notice that $\langle v_1,v_2 \rangle=|v_1||v_2|cos(\pi-A)=-cos(A)$,

$\langle v_1,v_3 \rangle=|v_1||v_3|cos(\pi-B)=-cos(B)$,

$\langle v_2,v_3 \rangle=|v_2||v_3|cos(\pi-C)=-cos(C)$.

Since every Gram matrix is positive semidefinite then $$(x,y,z)\begin{pmatrix}1&-\cos A&-\cos B\\-\cos A&1&-\cos C\\ -\cos B&-\cos C&1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}\geq 0.$$

Answer 2

The matrix is positive semidefinite. Its characteristic polynomial is $$ (1-X)^3-(1-X)(\cos^2A+\cos^2B+\cos^2C)-2\cos A\cos B\cos C. $$ If $A+B+C=\pi$, it's easy to prove that $$ 2\cos A\cos B\cos C=1-(\cos^2A+\cos^2B+\cos^2C) $$ (start from $\cos^2A+\cos^2B+\cos^2C$ and consider $C=\pi-(A+B)$). Set $k=\cos^2A+\cos^2B+\cos^2C$, so the characteristic polynomial can be written as $$ (1-X)^3-k(1-X)+k-1 $$ and, setting $t=1-X$, it becomes $t^3-kt+k-1$, which has $1$ as root and so it can be written as $$ (t-1)(t^2+t-k+1) $$ The root $t=1$ corresponds to a zero eigenvalue. The other eigenvalues are the roots of $$ (1-X)^2+(1-X)-k+1=X^2-3X+3-k=X^2-3X+(\sin^2A+\sin^2B+\sin^2C) $$ which has two positive roots.

Answer 3

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 $$x^2+y^2+z^2\geq2xy\cos{A}+2yz\cos{B}+2zx\cos{C}.$$