1/(p-a)^2 +1/(p - b)^2 +1/(p-c)^2 >=1 /r^2

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1/(p-a)^2 +1/(p - b)^2 +1/(p-c)^2 >=1 /r^2

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#1 h Oct 11, 2021, 6:20 AM

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Let $a, b$ , and $c$ be the lengths of the sides of a triangle and let $p$ the semiperimeter and $r$ the inradius of the triangle. Prove that
$\frac{1}{(p-a)^2}+\frac{1}{(p - b)^2}+\frac{1}{(p-c)^2}\ge \frac{1}{r^2}$

This post has been edited 1 time. Last edited by parmenides51, Oct 11, 2021, 6:28 AM

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#2 h Oct 11, 2021, 7:26 AM

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We know that $\left(\frac{r}{p-a}\right)^2=\left(\tan{\frac A2}\right)^2$ so it suffices to prove that $\left(\tan{\frac A2}\right)^2+\left(\tan{\frac B2}\right)^2+\left(\tan{\frac C2}\right)^2\ge 1$ .

$\left(\tan{\frac A2}\right)^2+\left(\tan{\frac B2}\right)^2+\left(\tan{\frac C2}\right)^2\ge \left(\tan{\frac A2}\right)\left(\tan{\frac B2}\right)+\left(\tan{\frac A2}\right)\left(\tan{\frac C2}\right)+\left(\tan{\frac B2}\right)\left(\tan{\frac C2}\right)=1$ , as desired.

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#3 h Oct 27, 2021, 3:52 AM

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BarisKoyuncu wrote:

We know that $\left(\frac{r}{p-a}\right)^2=\left(\tan{\frac A2}\right)^2$ so it suffices to prove that $\left(\tan{\frac A2}\right)^2+\left(\tan{\frac B2}\right)^2+\left(\tan{\frac C2}\right)^2\ge 1$ .

$\left(\tan{\frac A2}\right)^2+\left(\tan{\frac B2}\right)^2+\left(\tan{\frac C2}\right)^2\ge \left(\tan{\frac A2}\right)\left(\tan{\frac B2}\right)+\left(\tan{\frac A2}\right)\left(\tan{\frac C2}\right)+\left(\tan{\frac B2}\right)\left(\tan{\frac C2}\right)=1$ , as desired.

Can explain why $\left(\frac{r}{p-a}\right)^2=\left(\tan{\frac A2}\right)^2$ ?

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AlanLG

#4 h Oct 27, 2021, 4:14 AM

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Albert123 wrote:

BarisKoyuncu wrote:

We know that $\left(\frac{r}{p-a}\right)^2=\left(\tan{\frac A2}\right)^2$ so it suffices to prove that $\left(\tan{\frac A2}\right)^2+\left(\tan{\frac B2}\right)^2+\left(\tan{\frac C2}\right)^2\ge 1$ .

$\left(\tan{\frac A2}\right)^2+\left(\tan{\frac B2}\right)^2+\left(\tan{\frac C2}\right)^2\ge \left(\tan{\frac A2}\right)\left(\tan{\frac B2}\right)+\left(\tan{\frac A2}\right)\left(\tan{\frac C2}\right)+\left(\tan{\frac B2}\right)\left(\tan{\frac C2}\right)=1$ , as desired.

Can explain why $\left(\frac{r}{p-a}\right)^2=\left(\tan{\frac A2}\right)^2$ ?

in $\triangle API, \tan \frac{A}{2}=\frac{r}{AP}$ , thus you need to prove $AP=\frac{a+b+c}{2}-a$ , which is easy using that $AP=AR,BQ=BR,CQ=CP$

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#5 h Oct 27, 2021, 3:39 PM

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AlanLG wrote:

Albert123 wrote:

BarisKoyuncu wrote:

We know that $\left(\frac{r}{p-a}\right)^2=\left(\tan{\frac A2}\right)^2$ so it suffices to prove that $\left(\tan{\frac A2}\right)^2+\left(\tan{\frac B2}\right)^2+\left(\tan{\frac C2}\right)^2\ge 1$ .

$\left(\tan{\frac A2}\right)^2+\left(\tan{\frac B2}\right)^2+\left(\tan{\frac C2}\right)^2\ge \left(\tan{\frac A2}\right)\left(\tan{\frac B2}\right)+\left(\tan{\frac A2}\right)\left(\tan{\frac C2}\right)+\left(\tan{\frac B2}\right)\left(\tan{\frac C2}\right)=1$ , as desired.

Can explain why $\left(\frac{r}{p-a}\right)^2=\left(\tan{\frac A2}\right)^2$ ?

in $\triangle API, \tan \frac{A}{2}=\frac{r}{AP}$ , thus you need to prove $AP=\frac{a+b+c}{2}-a$ , which is easy using that $AP=AR,BQ=BR,CQ=CP$

Thank you

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