<s>[INST] If $aI am trying to prove $a^2<\frac{1}{3}(a^2+ab+b^2)<b^2$ if $a<b$. I am having trouble with the left inequality:
$a^2<\frac{1}{3}(a^2+ab+b^2) \Rightarrow 2a^2<ab+b^2$.
If $a>0$, then all is well and adding $a^2<b^2$ to $a^2<ab$ will yield the result I want, but I do not know how to deal with the case $a<0$, or if there is any simpler ways of proving the inequality.
 [/INST] Is true only if $0< a<b$, then for the left side:
$$a^2<b^2
$$
 $$a^2+a^2+ab<b^2+ab+a^2
$$
$$a^2+a^2+ab<b^2+ab+a^2
$$
If $a<b$ then how $a>0$ then $a^2<ab$
$$a^2+a^2+a^2<a^2+a^2+ab<b^2+ab+a^2
$$
$$3a^2 <b^2+ab+a^2
$$
For the rigth side
$$a^2<b^2
$$
 $$a^2+b^2+ab<b^2+b^2+ab
$$
If $a<b$ then how $b>0$ then $ab<b^2$
 $$a^2+ab+b^2<b^2+b^2+ab<b^2+b^2+b^2
$$
$$a^2+ab+b^2<3b^2
$$
</s>