Fie \( ABC \) un triunghi echilateral de latura 1 si fie \( M\in\left[AB\right],\ N\in\left[BC\right],\ P\in\left[CA\right] \) astfel incat \( AM+BN+CP\ge2. \). Aratati ca:
a) Pentru orice numere reale \( x,\ y,\ z\in\left[0,\ 1\right] \) avem: \( \frac{1}{xy+2}+\frac{1}{yz+2}+\frac{1}{zx+2}\le\frac{3}{x+y+z} \).
b) \( 1\le\frac{1}{AM\cdot BN+2}+\frac{1}{BN\cdot CP+2}+\frac{1}{CP\cdot AM+2}\le\frac{3}{2} \)
c) \( MN^{2}+NP^{2}+PM^{2}\ge\frac{4\sqrt{3}}{3}\cdot S_{MNP}+\frac{1}{2} \)
Claudiu Mindrila, R. M. T. 1/2010
Inegalitate in triunghi echilateral
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Claudiu Mindrila
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Inegalitate in triunghi echilateral
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Andi Brojbeanu
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a) \( x,y,z\in [0; 1] \Rightarrow (x-1)(y-1)\ge 0 \) si analoagele \( \Rightarrow xy-x-y+1\ge 0\Rightarrow xy+2\ge x+y+1 \) si analoagele.
\( LHS=\sum {\frac{1}{xy+2}}\le \sum {\frac{1}{x+y+1}}\le \sum{\frac{1}{x+y+z}}=\frac{3}{x+y+z}=RHS \).
b) Observam ca \( AM, BN, CP\in [0;1] \) si din a)\( \Rightarrow \sum {\frac{1}{AM\cdot BN+2}}\le \frac{3}{AM\cdot BN+BN\cdot CP+ CP\cdot AM}\le \frac{3}{2} \).
\( \sum {\frac{1}{AM\cdot BN+2}}\ge \sum{\frac{1}{1\cdot 1+2}}=\sum{\frac{1}{3}}=1 \).
c)\( MN^2=BM^2+BN^2-2\cdot BM\cdot BN\cdot cos(ABC)=BM^2+BN^2-BM\cdot BN \) si analoagele.
\( S_{MNP}=S_{ABC}-(S_{BMN}+S_{PNC}+S_{AMP})=S_{ABC}-\sum{\frac{BM\cdot BN\cdot sin(ABC)}{2}}=\frac{\sqrt{3}\cdot 1}{4}-\frac{\frac{\sqrt{3}}{2}}{2}(\sum {BM\cdot BN})=\frac{\sqrt{3}}{4}(1-\sum BM\cdot BN)\Leftrightarrow \sum{AM^2}\ge\frac{3}{2} \)
\( LHS\ge RHS\Leftrightarrow AM^2+BM^2+BN^2+NC^2+PC^2+AP^2-(BM\cdot BN+NC\cdot PC+AP\cdot AM)\ge \frac{4}{\sqrt{3}}\cdot \frac{\sqrt{3}}{4}[1-(BM\cdot BN+CN\cdot CP+AP\cdot AM]+\frac{1}{2}\Leftrightarrow \)
\( \Leftrightarrow \sum{AM^2}-\sum {BM\cdot BN}\ge 1-\sum{BM\cdot BN}+\frac{1}{2}\Leftrightarrow \sum{AM^2}\ge\frac{3}{2} \).
Ultima inegalitate este evidenta din inegalitatea \( C.B.S.:\sum{AM^2}\ge\frac{(AM+BM+BN+CN+CP+AP)^2}{6}=\frac{3^2}{6}=\frac{3}{2} \).
\( LHS=\sum {\frac{1}{xy+2}}\le \sum {\frac{1}{x+y+1}}\le \sum{\frac{1}{x+y+z}}=\frac{3}{x+y+z}=RHS \).
b) Observam ca \( AM, BN, CP\in [0;1] \) si din a)\( \Rightarrow \sum {\frac{1}{AM\cdot BN+2}}\le \frac{3}{AM\cdot BN+BN\cdot CP+ CP\cdot AM}\le \frac{3}{2} \).
\( \sum {\frac{1}{AM\cdot BN+2}}\ge \sum{\frac{1}{1\cdot 1+2}}=\sum{\frac{1}{3}}=1 \).
c)\( MN^2=BM^2+BN^2-2\cdot BM\cdot BN\cdot cos(ABC)=BM^2+BN^2-BM\cdot BN \) si analoagele.
\( S_{MNP}=S_{ABC}-(S_{BMN}+S_{PNC}+S_{AMP})=S_{ABC}-\sum{\frac{BM\cdot BN\cdot sin(ABC)}{2}}=\frac{\sqrt{3}\cdot 1}{4}-\frac{\frac{\sqrt{3}}{2}}{2}(\sum {BM\cdot BN})=\frac{\sqrt{3}}{4}(1-\sum BM\cdot BN)\Leftrightarrow \sum{AM^2}\ge\frac{3}{2} \)
\( LHS\ge RHS\Leftrightarrow AM^2+BM^2+BN^2+NC^2+PC^2+AP^2-(BM\cdot BN+NC\cdot PC+AP\cdot AM)\ge \frac{4}{\sqrt{3}}\cdot \frac{\sqrt{3}}{4}[1-(BM\cdot BN+CN\cdot CP+AP\cdot AM]+\frac{1}{2}\Leftrightarrow \)
\( \Leftrightarrow \sum{AM^2}-\sum {BM\cdot BN}\ge 1-\sum{BM\cdot BN}+\frac{1}{2}\Leftrightarrow \sum{AM^2}\ge\frac{3}{2} \).
Ultima inegalitate este evidenta din inegalitatea \( C.B.S.:\sum{AM^2}\ge\frac{(AM+BM+BN+CN+CP+AP)^2}{6}=\frac{3^2}{6}=\frac{3}{2} \).