If $a<b$, then $a^2<\frac{1}{3}(a^2+ab+b^2)<b^2$ I am trying to prove $a^2&lt;\frac{1}{3}(a^2+ab+b^2)&lt;b^2$ if $a&lt;b$. I am having trouble with the left inequality:
$a^2&lt;\frac{1}{3}(a^2+ab+b^2) \Rightarrow 2a^2&lt;ab+b^2$.
If $a&gt;0$, then all is well and adding $a^2&lt;b^2$ to $a^2&lt;ab$ will yield the result I want, but I do not know how to deal with the case $a&lt;0$, or if there is any simpler ways of proving the inequality.
The inequality doesn't generally hold if $a&lt;0$ so you don't need to worry about it. e.g. consider $a=-1$ and $b=1$.
Is true only if $0&lt; a&lt;b$, then for the left side:

$$a^2&lt;b^2
$$
 $$a^2+a^2+ab&lt;b^2+ab+a^2
$$
$$a^2+a^2+ab&lt;b^2+ab+a^2
$$
If $a&lt;b$ then how $a&gt;0$ then $a^2&lt;ab$
$$a^2+a^2+a^2&lt;a^2+a^2+ab&lt;b^2+ab+a^2
$$
$$3a^2 &lt;b^2+ab+a^2
$$

For the rigth side

$$a^2&lt;b^2
$$
 $$a^2+b^2+ab&lt;b^2+b^2+ab
$$

If $a&lt;b$ then how $b&gt;0$ then $ab&lt;b^2$
 $$a^2+ab+b^2&lt;b^2+b^2+ab&lt;b^2+b^2+b^2
$$
$$a^2+ab+b^2&lt;3b^2
$$