Doua inegalitati (nu prea simple !) intr-un triunghi (own).

[Virgil Nicula]

Sa de arate ca intr-un triunghi \( ABC \) (notatii standard) exista relatiile :

\( 1.\ \ \underline{\overline{\left\|\ \sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot (a+b-c)}\ \le\ \frac {ab+bc+ca}{a+b+c}\ \right\|}}\ \le\ \frac {a+b+c}{3}\ . \)

\( 2.\ \ \ 3\ \le\ \underline{\overline{\left\|\ \frac {(a+b+c)^2}{ab+bc+ca}\ \le\ \sqrt 3\cdot\sqrt[3]{\frac pr}\ \right\|}}\ . \)

[maxim bogdan]

Sa de arate ca intr-un triunghi \( ABC \) (notatii standard) exista relatiile :

\( 1.\ \ \underline{\overline{\left\|\ \sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot (a+b-c)}\ \le\ \frac {ab+bc+ca}{a+b+c}\ \right\|}}\ \le\ \frac {a+b+c}{3}\ . \)

Solutie. Deconditionam inegalitatea cu substitutiile: \( a\to b^{3}+c^{3}, b\to c^{3}+a^{3}, c\to a^{3}+b^{3}(a,b,c\in\mathbb{R}_{+}). \)

Inegalitatea va fi echivalenta cu: \( 2abc\leq\frac{\sum a^{6}+3\sum a^3 b^3}{2(\sum a^3)}\Longleftrightarrow \underline{\overline{\left\| F(a,b,c)= \sum a^6 +3\sum a^3 b^3 - 4\sum a^4 bc \geq 0 \right\|}}. \)

Ne propunem sa scriem \( F(a,b,c) \) sub forma \( F(a,b,c)=S_{c}(a-b)^2+S_{a}(b-c)^2+S_{b}(c-a)^2, \) unde \( S_{a},S_{b}, S_{c} \) sunt expresii in functie de \( a,b \) si \( c. \) (stim cu siguranta ca exista din teorema de reprezentare simetrica SOS deoarece \( F(a,b,c) \) e un polinom simetric standard).

Dupa ceva timp am gasit ca: \( \left \{\begin{array}{rcl} S_{a}=4a^4 +b^4 +c^4 +2b^3 c+2c^3 b-b^2 c^2\geq 0\\ \\ S_{b}=a^4 +4b^4 +c^4 +2a^3 c+2 c^3 a-a^2 c^2\geq 0\\ \\ S_{c}=a^4+b^4+4c^4 +2a^3 b+2b^3 a-a^2 b^2\geq 0 \end{array}\right. \)

Deci din "Teorema SOS"(SOS Theorem): \( F(a,b,c)=\sum (a^4 +b^4 +4c^4+2a^3 b+2b^3 a-a^2 b^2)(a-b)^2\geq 0. \)