Inegalitate remarcabila in triunghi

[opincariumihai]

Aratati ca in orice triunghi are loc inegalitatea : \( HG \geq IH \) .
Deduceti folosind aceasta inegalitate ca :
1. \( OI \geq OG \) ( Balcaniada 96 ) ;
2. Inegalitatea propusa aici .

[Mateescu Constantin]

Aratati ca in orice triunghi are loc inegalitatea : \( HG \geq IH \) .

Folosim formulele: \( \left\|\begin{array}{cc} HG^2 & = & 4R^2-\frac 49 (a^2+b^2+c^2) \\\\\\\
IH^2 & = & 4R(R+r)+3r^2-p^2 \end{array}\right\| \)

\( \Longrightarrow\ HG\ge IH\ \Longleftrightarrow\ 4R^2-\frac 49(a^2+b^2+c^2)\ \ge\ 4R(R+r)+3r^2-p^2 \)

\( \Longleftrightarrow\ 4R^2-\frac 49(2p^2-8Rr-2r^2)\ \ge\ 4R^2+4Rr+3r^2-p^2\ \Longleftrightarrow\ p^2\ \ge\ 19r^2+4Rr\ (*) \)

Pe de alta parte \( GI^2=\frac{p^2+5r^2-16Rr}{9}\ \ge\ 0 \), deci \( p^2\ \ge\ 16Rr-5r^2 \)

Ramane sa aratam ca \( 16Rr-5r^2\ \ge\ 19r^2+4Rr\ \Longleftrightarrow\ R\ge 2r \), care este adevarata. Deci \( (*) \) este indeplinita.

[Mateescu Constantin]

\( OI \geq OG \) (Balcaniada 1996)

Tinem cont de: \( \left\|\begin{array}{cc} OI^2 & = & R^2-2Rr \\\\\\\\\\
OG^2 & = & R^2-\frac 19(a^2+b^2+c^2)\end{array}\right\| \) .

Astfel ramane sa demonstram ca: \( a^2+b^2+c^2\ \ge\ 18Rr \) .

\( \Longleftrightarrow\ a^2+b^2+c^2\ \ge\ \frac{9abc}{a+b+c}\ \Longleftrightarrow\ (a+b+c)(a^2+b^2+c^2)\ \ge\ 9abc=3\sqrt[3]{abc}\ \cdot\ 3 \sqrt[3]{a^2b^2c^2} \) , evident.

[Mateescu Constantin]

\( OI+OG\ \ge\ IH \)

Folosind inegalitatea \( OI\ \ge\ OG \) ramane sa aratam numai ca \( 2OG\ \ge\ IH\ \Longleftrightarrow\ HG\ \ge\ IH \), adevarat.