First, we replace $$a=y+z, b=x+z, c=x+y$$ with $$x, y, z > 0$$ being tangents on the inscribed circle, as in

![trick](https://i.sstatic.net/IQ0ve.png)

The inequality thus becomes

$$
\frac{1}{3} \leq \frac{(y+z)^2 + (x+z)^2 + (x+y)^2}{(y+z+x+z+x+y)^2} < \frac{1}{2}
$$

which is

$$
\frac{1}{3} \leq \frac{(x^2+y^2+z^2)+(xy+xz+yz)}{2(x^2+y^2+z^2)+4(xy+xz+yz)} < \frac{1}{2}
$$

or

$$
\frac{1}{3} \leq \frac{u+v}{2u+4v} < \frac{1}{2}
$$

First, for the upper bound

$$
\frac{u+v}{2u+4v} < \frac{1}{2}
$$

$$
2u+2v < 2u+4v
$$

$$
v > 0
$$

which is obviously true because $$v$$ is

$$
xy + yz+ xz > 0
$$

Second, for the lower bound

$$
\frac{u+v}{2u+4v} \geq \frac{1}{3}
$$

$$
3u+3v \geq 2u+4v
$$

$$
u \geq v
$$

Is equivalent to

$$
x^2+y^2+z^2 \geq xy+yz+xz
$$

$$
x^2+y^2+z^2-xy-xz-yz \geq 0
$$

$$
(x-y)^2 + (x-z)^2 + (y-z)^2 >= 0
$$

The sum must be >= 0 because each summand is.

Scroogey

The inequality with 1/3 is a special case of the [Cauchy inequality](https://en.m.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality) 😉

oliverweiss

42. 

Here's proof

https://en.m.wikipedia.org/wiki/42_(number)#:~:text=The%20number%2042%20is%2C%20in,knows%20what%20the%20question%20is.

"Answer to the Ultimate Question of Life, the Universe, and Everything"

Aardvark

science

This inequality holds for any triangle:
$$
\frac{1}{3} \leq \frac{a^2 + b^2 + c^2}{(a + b + c)^2} < \frac{1}{2}
$$
where $$a$$, $$b$$ and $$c$$ correspond to the lengths of the triangle.

Can you prove it?

Previous iteration: https://stacker.news/items/760106/r/south_korea_ln (brief and corrected answer in https://stacker.news/items/760106/r/south_korea_ln?commentId=760229; I will write out a more detailed version of this same answer as well as an alternative approach using the tau function or divisor function, _when I am a bit less tired_)