Inegalitate in orice triunghi >=7

[Cezar Lupu]

Sa se arate ca in orice triunghi avand laturile de lungimi \( a, b, c \) este adevarata inegalitatea:

\( \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}+\frac{(b+c-a)(c+a-b)(a+b-c)}{abc}\geq 7 \).

Cezar Lupu, Mathematical Reflections 2009

[Tudor Micu]

Fie \( a=y+z \) \( b=z+x \) \( c=x+y \)
Inegalitatea revine la
\( \sum_{cyc}\frac{2x+y+z}{y+z}+\frac{8xyz}{(x+y)(y+z)(z+x)}\geq 7 \)
\( \sum_{cyc}\frac{2x}{y+z}+\frac{8xyz}{(x+y)(y+z)(z+x)}\geq 4 \)
\( \sum_{cyc}(x(x+y)(z+x))+4xyz\geq 2(x+y)(y+z)(z+x) \)
\( \sum_{cyc}x^3+\sum_{cyc}x^2y+\sum_{cyc}xy^2+7xyz\geq 2\sum_{cyc}x^2y+2\sum_{cyc}xy^2+4xyz \)
\( x^3+y^3+z^3+3xyz\geq x^2y+y^2z+z^2x+xy^2+yz^2+zx^2 \),
care rezulta din inegalitatea lui Schur.

[Theodor Munteanu]

O inegalitate ceva mai usoara.

\( \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}+\frac{abc}{(b+c-a)(c+a-b)(a+b-c)}\geq 7 \).

[Virgil Nicula]

Sa se arate ca in orice triunghi avand laturile de lungimi \( a, b, c \) este adevarata inegalitatea:

\( \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}+\frac{(b+c-a)(c+a-b)(a+b-c)}{abc}\ \ge\ 7 \) (Cezar Lupu, Mathematical Reflections 2009).

Dem. \( \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}+\frac{(b+c-a)(c+a-b)(a+b-c)}{abc}\ \ge\ 7\ \Longleftrightarrow\ \sum a\cdot\sum \frac 1a+\frac {8pr^2}{4Rpr}\ \ge\ 10\ \Longleftrightarrow \)

\( \frac {p^2+r^2+4Rr}{2Rr}+\frac {2r}{R}\ \ge\ 10\ \Longleftrightarrow\ \underline{\overline{\left\|\ p^2+5r^2\ \ge\ 16Rr\ \right\|}} \) care se stie sau se arata \( ^{(*)} \) usor.

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\( ^{(*)}\ \) LEMA. \( \{x\ ,\ y\ ,\ z\}\ \subset\ R\ \wedge\ x+y+z=0\ \Longrightarrow\ yza^2+zxb^2+xyc^2\ \le\ 0\ . \)

Dem. \( \left\|\ \begin{array}{c}
yza^2+zxb^2+xyc^2\ \le\ 0\\\\
z=-(x+y)\end{array}\ \right\|\ \Longrightarrow\ yza^2\le (y+z)\left(zb^2+yc^2\right)\ \Longleftrightarrow \)

\( y^2c^2+yz\left(b^2+c^2-a^2\right)+z^2b^2\ge 0\ \Longleftrightarrow\ y^2c^2+2yzbc\cdot \cos A+z^2b^2\ge 0\ \Longleftrightarrow \)

\( \left(yc+zb\cdot\cos A\right)^2+z^2b^2\sin^2A\ge 0\ , \) ceea ce este adevarat. Avem egalitate \( \Longleftrightarrow\ x=y=z=0 . \)

Caz particular. \( \left|\ \begin{array}{c}
x=b+c-2a\\\\\\
y=c+a-2b\\\\\\
z=a+b-2c\end{array}\ \right|\Longrightarrow\ \sum (c+a-2b)(a+b-2c)a^2\le0\ \Longleftrightarrow s_1^3-5s_1s_2+18s_3\le 0\Longleftrightarrow \)

\( 8p^3-10p\left(p^2+r^2+4Rr\right)+18\cdot 4Rpr\le 0\Longleftrightarrow\underline{\overline{\left\|\ p^2+5r^2\ \ge\ 16Rr\ \right\|}}\ . \) Acestei binecunoscute

inegalitati i se poate da si o interpretare geometrica, anume \( \underline{\overline{\left\|\ 9\cdot GI^2=p^2+5r^2-16Rr\ \ge\ 0\ \right\|}}\ . \) Intr-adevar,

folosind identitatea Leibniz \( \sum XA^2=3\cdot XG^2+\frac 13\cdot\sum a^2 \) pentru \( X:=I \) si stiind ca \( IA^2=\frac {bc(p-a)}{p} \)

etc se obtine : \( 3\left(p^2+r^2-8Rr\right)=9\cdot IG^2+2\left(p^2-r^2-4Rr\right)\Longleftrightarrow 9\cdot IG^2=p^2+5r^2-16Rr\ . \)