<s>[INST] Inequality in a triangle associated with Golden ratio 
Let $ABC$ be arbitrary triangle, $D$, $E$, $F$ are the midpoints of $BC$, $CA$, $AB$ respectively. Define points, segments in the figure below. I am looking for a proof that:

$$DE+EF+FD \le (DG+DH+EI+EJ+FJ+FQ).\frac{\varphi}{2}$$

Wher $\varphi=\frac{\sqrt{5}+1}{2}$ the golden ratio.


Equality if only if $ABC$ is the equilateral triangle.

See also:

*

*Golden ratio as a property of conic section (is it known?)

 [/INST] Let $DH=x$, $DG=y$, $FK=z$, $FQ=t$, $EG=u$ and $EI=v$.
Thus, in the standard notation we obtain:
$$u\left(y+\frac{c}{2}\right)=\frac{b^2}{4},$$$$y\left(u+\frac{c}{2}\right)=\frac{a^2}{4},$$
Which gives $$u-y=\frac{b^2-a^2}{2c},$$
$$y\left(y+\frac{b^2-a^2}{2c}\right)+\frac{c}{2}y=\frac{a^2}{4},$$
$$y=-\frac{b^2+c^2-a^2}{4c}+\sqrt{\left(\frac{b^2+c^2-a^2}{4c}\right)^2+\frac{a^2}{4}},$$
$$u=-\frac{a^2+c^2-b^2}{4c}+\sqrt{\left(\frac{b^2+c^2-a^2}{4c}\right)^2+\frac{a^2}{4}}$$ and by C-S we have:
$$u+y=-\frac{c}{2}+\sqrt{\left(\frac{b^2+c^2-a^2}{2c}\right)^2+a^2}=$$
$$=-\frac{c}{2}+\sqrt{\frac{(a^2-b^2)^2+c^2(c^2+2a^2+2b^2)}{4c^2}}\geq-\frac{c}{2}+\frac{1}{2}\sqrt{c^2+2a^2+2b^2}=$$$$=-\frac{c}{2}+\frac{1}{2\sqrt5}\sqrt{(1+2+2)(c^2+2a^2+2b^2)}\geq-\frac{c}{2}+\frac{c+2a+2b}{2\sqrt5}.$$
By the same way we obtain:
$$v+t\geq-\frac{a}{2}+\frac{a+2b+2c}{2\sqrt5}$$ and
$$x+z\geq-\frac{b}{2}+\frac{b+2a+2c}{2\sqrt5}.$$
Can you end it now?
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