Let $BE$ and $CF$ be the bisectors of a triangle $ABC$. Prove that $2EF \leq BF + CE$.
Problem
Source: 10.3 of XX Geometrical Olympiad in honour of I.F.Sharygin
Tags: geo, geometry, Inequality
NO_SQUARES
07.08.2024 09:52
In fact, $BF\cdot CE \geqslant EF^2$ is also true (and seems easier). We will prove exactly this inequality.
Let $a=BC, b=CA, c=AB$, so \[AF=\frac{bc}{a+b}, AE=\frac{bc}{a+c}, BF=\frac{ac}{a+b}, CE=\frac{ab}{a+c} \]and so by cosine theorem \[ EF^2=AF^2+AE^2-2AF \cdot AE \cdot cos(A) =bc\left( \frac{bc}{(a+b)^2}+\frac{bc}{(a+c)^2}-\frac{b^2+c^2-a^2}{(a+b)(a+c)} \right ), \]thus our inequality, after expanding and dividing on $abc$, is equivalent to \[2abc+b^2c+bc^2 \leqslant a(b^2+c^2)+(b^3+c^3) \]which is true since $b^3+c^3 \geqslant b^2c+bc^2$ and $a(b^2+c^2) \geqslant 2abc$. $\square$
sami1618
14.08.2024 17:06
Made mistake
NO_SQUARES
14.08.2024 23:10
sami1618 wrote:
$$2EF\leq 2(IF+IE)\leq BF+CE$$
The second one is wrong, of course. Try $AB=AC$.
buratinogigle
15.08.2024 04:39
NO_SQUARES wrote:
In fact, $BF\cdot CE \geqslant EF^2$ is also true (and seems easier). We will prove exactly this inequality.
Let $a=BC, b=CA, c=AB$, so \[AF=\frac{bc}{a+b}, AE=\frac{bc}{a+c}, BF=\frac{ac}{a+b}, CE=\frac{ab}{a+c} \]and so by cosine theorem \[ EF^2=AF^2+AE^2-2AF \cdot AE \cdot cos(A) =bc\left( \frac{bc}{(a+b)^2}+\frac{bc}{(a+c)^2}-\frac{b^2+c^2-a^2}{(a+b)(a+c)} \right ), \]thus our inequality, after expanding and dividing on $abc$, is equivalent to \[2abc+b^2c+bc^2 \leqslant a(b^2+c^2)+(b^3+c^3) \]which is true since $b^3+c^3 \geqslant b^2c+bc^2$ and $a(b^2+c^2) \geqslant 2abc$. $\square$
I think that your way of strengthening the inequality is very interesting. My students were unable to solve this problem using pure geometry during the exam period. An interesting consequence of this inequality is that the perimeter of triangle $DEF$ does not exceed the semi-perimeter of triangle $ABC$. This is also a pretty strong inequality.
kiyoras_2001
15.08.2024 11:42
buratinogigle wrote: An interesting consequence of this inequality is that the perimeter of triangle $DEF$ does not exceed the semi-perimeter of triangle $ABC$. This inequality is known. For example see here.