On the sides $BC, CA$ and $AB$ of triangle $ABC$, respectively, points $D, E$ and $F$ are chosen. Prove that $\frac12 (BC + CA + AB)<AD + BE + CF<\frac 32 (BC + CA + AB)$.
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Tags: geometry, geometric inequality
SlimTune
16.10.2020 19:51
I think I found a solution for the upper bound.
Let the sides of $\triangle{ABC}$ be denoted by $x, y, z$ where $x \geq y \geq z$.
Lemma 1: The largest possible value of $AD + BE + CF$ is $2x + y$.
ProofThe side with length $x$ has two endpoints $A$ and $B$ (WLOG) which are vertices of $\triangle ABC$. Since $x$ is greater than the other two side lengths, this means that the maximum value of $AD$ and $BE$ is $x$ for each of them.
Now, we will address $C$. The two sides with $C$ have length $y$ and $z$. Since $y$ is larger than $z$, we have that the maximum value of $CF$ is $y$.
Hence, the max value of $AD + BE + CF$ is $2x+y$.
Now we need to prove that $2x + y < \frac{3}{2}(x + y + z)$. We can expand and rearrange the inequality to get $x < 3z + y$. We see that this is true from the Triangle Inequality ($x < z + y$), so we are done. $\square$
Please correct me if my proof is incorrect.
natmath
16.10.2020 21:50
@above your proof is correct. $AD$, $BE$, and $CF$ are all independent of eachother. $AD\geq h_a$ where $h_a$ is the altitude of $A$. $a*h_a=\frac{1}{2}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}$ Now we need to prove $a+b+c<\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$