[Daily puzzle] Triangle inequality \ stacker news ~science

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550 sats \ 3 comments \ @south_korea_ln 10 Nov 2024 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: #760106 (brief and corrected answer in #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)

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501 sats \ 0 replies \ @Scroogey 10 Nov 2024

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

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

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

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

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

 $xy + yz+ xz > 0$

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.

147 sats \ 0 replies \ @oliverweiss 10 Nov 2024

The inequality with 1/3 is a special case of the Cauchy inequality 😉

42 sats \ 0 replies \ @Aardvark 10 Nov 2024

Here's proof

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

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