<s>[INST] Are these metrics? I want to find if the below functions are metrics. I have worked through each of the three conditions, but am stuck on the positivity of $f(a, b)$ (first condition-see below) and the triangle inequality of $g(a, b)$ (third condition-see below). 

Edit: Have now solves these two parts, see the answers for positivity of $f(a, b)$  and the triangle inequality of $g(a, b)$

Please could you help me on these two points?

$f(a, b)=a^2+ab+b^2$



*

*Initially I thought this would not satisfy positivity. But for $f$
to be negative, $ab$ must be negative, and greater in modulus value
to $a^2+b^2$. So for this to work, $a$ and $b$ must be opposite sign
and less than $1$. I have tried small opposite signed values of $a$
and $b$ but have not found $f(a, b)$ to be positivity. Not sure.

*Symmetry. $f(b, a)=b^2+ba+a^2=a^2+ab+b^2$ so this is satisfied

*Triangle inequality. Is $f(a, b)+f(b, c) \geq f(a, c)$ satisfied? Let $X=f(a, b)+f(b, c)-f(a, c)=a^2+b^2+ab+c^2+b^2+cb-a^2-c^2-ac=b^2+ab+cb-ac$. Assume $X \geq 0 $ for all $a, b, c \in \mathbb{R}\implies b^2 +ab+cb \geq ac$. My initial thought is that if we take $b$ to be very small and $a$ and $c$ to be comparatively larger, then we may arrive at a contradiction. Take $b=0.01, a=100, c=100 \implies 0.0001+2 \geq 10000$. Contradiction. Triangle inequality does not hold.



$g(a, b)= \frac{|a^3-b^3|}{2}$ 



*

*Positivity satisfied since $|x| \geq 0 \forall x \in \mathbb{R}$ and $2 \geq 0$

*$g(b, a)=\frac{|b^3-a^3|}{2}=\frac{|-(-b^3+a^3)|}{2}=\frac{|-1||a^3-b^3|}{2}=\frac{|a^3-b^3|}{2}=g(a, b)$. So symmetry is satisfied.

*Must consider the sign of: $X=g(a, b)+g(b, c)-g(a, c)=\frac{|a^3-b^3|}{2}+\frac{|b^3-c^3|}{2}-\frac{|a^3-c^3|}{2}$ For the triangle inequality to be satisfied $X$ must be positive. I tried to find some counter-examples, such as take $a=100, c=0 \implies \frac{|100^3-b^3|}{2}+\frac{|b^3|}{2}-\frac{|100^3|}{2}$. But if $b$ is positive then the whole thing is positive. If it is negative , then the first two terms are positive and larger than the third. From this reasoning, I believe $X$ is always positive and the triangle inequality holds.  Not sure.
 [/INST] So $$|x|+|y|\ge |x+y|,$$ then
$$\frac{|a^3-b^3|}{2}+\frac{|b^3-c^3|}{2}=\frac{|a^3-b^3|+|b^3-c^3|}{2}\ge$$
$$\ge \frac{|a^3-b^3+b^3-c^3|}{2}=\frac{|a^3-c^3|}{2}$$
</s>