Inegalitate frumoasa

[Theodor Munteanu]

\( 9a^2 b^2 c^2 + (a^2 + b^2 + c^2 )(a^4 + b^4 + c^4 - 2a^2 b^2 - 2b^2 c^2 - 2a^2 c^2 ) > 0 \), unde a, b, c lungimile laturilor intr-un triunghi.

[opincariumihai]

Cum \( 16S^2=2\sum a^2b^2-\sum c^4 \) inegalitatea de dem. se scrie echivalent \( 9(abc)^2\geq16(a^2+b^2+c^2)S^2 \) sau \( 9R^2\geq a^2+b^2+c^2 \) inegalitate adevarata.
Egalitatea are loc pentru \( a=b=c \)

[Theodor Munteanu]

\(
\begin{array}{l}
\left. {\left\| {a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)} \right.} \right\| \\
{\rm (a}^{\rm 2} + b^2 + c^2 )(a^4 + b^4 + c^4 - a^2 b^2 - b^2 c^2 - a^2 c^2 ) - {\rm (a}^{\rm 2} + b^2 + c^2 )(a^2 b^2 + b^2 c^2 + a^2 c^2 ) + 9a^2 b^2 c^2 > 0 \Leftrightarrow a^6 + b^6 + c^6 - 3a^2 b^2 c^2 - ({\rm a}^{\rm 2} + b^2 + c^2 )(a^2 b^2 + b^2 c^2 + a^2 c^2 ) + \\
9a^2 b^2 c^2 > 0 \Leftrightarrow a^6 + b^6 + c^6 + 6a^2 b^2 c^2 > ({\rm a}^{\rm 2} + b^2 + c^2 )(a^2 b^2 + b^2 c^2 + a^2 c^2 ) \Leftrightarrow x^3 + y^3 + z^3 + 6xyz > (x + y + z)(xy + yz + zx) \Leftrightarrow \\
x^3 + y^3 + z^3 + 6xyz > 3xyz + xy(x + y) + yz(y + z) + xz(x + z)|| - 3xyz \Leftrightarrow x^3 + y^3 + z^3 + 3xyz > xy(x + y) + yz(y + z) + xz(x + z); \\
x(x - y)(x - z) + y(y - x)(y - z) + z(z - x)(z - y) \ge 0(Schur,r = 1) \Leftrightarrow x(x^2 - xy - xz + yz) + y(y^2 - yx - yz + zx) + z(z^2 - zx - zy + xy) \ge 0 \Leftrightarrow \\
x^3 + y^3 + z^3 + 3xyz \ge xy(x + y) + yz(y + z) + zx(z + x). \\
\end{array}
\)

[opincariumihai]

Nu inteleg de ce sa facem atatea calcule algebrice , cand de fapt e vorba despre o inegalitate in triunghi si inca una destul de slaba