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Sa se arate ca intr-un triunghi \( ABC \) exista lantul de inegalitati

\( a^3+b^3+c^3-3abc\ \ge\ 2\cdot\left[abc-(b+c-a)(c+a-b)(a+b-c)\right]\ \ge\ 0 \) .

Virgil Nicula wrote:Sa se arate ca intr-un triunghi \( ABC \) exista lantul de inegalitati

\( a^3+b^3+c^3-3abc\ \ge\ 2\cdot\left[abc-(b+c-a)(c+a-b)(a+b-c)\right] \) .

\( \Leftrightarrow \frac{1}{2}\sum_{cyc}(a-b)^2(a+b+c)\ge \sum_{cyc}(a+b-c)(a-b)^2\Leftrightarrow \frac{1}{2}\sum_{cyc}(a-b)^2(3c-a-b)\ge 0 \)

WLOG \( a\ge b\ge c \). Atunci \( (c-a)^2\ge (a-b)^2+(b-c)^2 \) (deoarece \( \Leftrightarrow (a-b)(b-c)\ge 0 \)) si \( 3b-c-a\ge 0 \) (deoarece \( \Leftrightarrow (b+c-a)+2(b-c)\ge 0 \))

\( (a-b)^2(3c-a-b)+(b-c)^2(3a-b-c)+(c-a)^2(3b-c-a)\ge (a-b)^2(3c-a-b)+(b-c)^2(3a-b-c)+ \)

\( +(a-b)^2(3b-c-a)+(b-c)^2(3a-b-c)=2(a-b)^2(b+c-a)+2(b-c)^2(a+b-c)\ge 0 \)

Virgil Nicula wrote:

\( a^3+b^3+c^3-3abc \ge 2\cdot abc-(b+c-a)(c+a-b)(a+b-c) \)

Notam \( x=b+c-a \) si analoagele ,

\( \sum\(\frac{x+y}{2}\)^3+2xyz\ge \frac{5}{8}(x+y)(y+z)(z+x) \) sau

\( x^3+y^3+z^3+3xyz\ge x^2(y+z)+y^2(z+x)+z^2(x+y) \)

Virgil Nicula wrote:Sa se arate ca intr-un triunghi \( ABC \) exista lantul de inegalitati

\( a^3+b^3+c^3-3abc\ \ge\ 2\cdot\left[abc-(b+c-a)(c+a-b)(a+b-c)\right]\ \ge\ 0 \) .

Se stie sau se arata usor ca : \( \left\|\ \begin{array}{cc}
a^3+b^3+c^3=2p(p^2-6Rr-3r^2) \\\\\\\\\\\
\prod(b+c-a)=8\prod(p-a)=8pr^2 \ \end{array}\right\| \) .

Inegalitatea din stanga este echivalenta cu : \( a^3+b^3+c^3+2\prod(b+c-a)\ \ge\ 5abc \)

\( \Longleftrightarrow\ 2p(p^2-6Rr-3r^2)+16pr^2\ \ge\ 20Rrp\ \Longleftrightarrow\ p^2-6Rr-3r^2+8r^2\ \ge\ 10Rr \)

\( \Longleftrightarrow\ p^2+5r^2\ \ge\ 16Rr\ ,\ \mbox{O.K.} \) .

In fine , partea dreapta se reduce la \( 4Rrp\ \ge\ 8pr^2\ \Longleftrightarrow\ R\ \ge\ 2r\ ,\ \mbox{O.K.} \)