Fie \( a,b,c \) numere pozitive. Sa se demonstreze ca
\( (\frac{a}{a+2b})^2+(\frac{b}{b+2c})^2+(\frac{c}{c+2a})^2 \geq \frac{1}{3} \)
Inegalitate ciclica-easy
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Claudiu Mindrila
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Omer Cerrahoglu
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Este suficient sa aratam ca \( \frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2a}\geq{1} \)(din \( MA\geq MP \) avem inegalitatea data).
Din CBS avem ca \( (\frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2a})[a(a+2b)+b(b+2c)+c(c+2a)]\geq (a+b+c)^2 \Longleftrightarrow \frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2a}\geq \frac{a^2+b^2+c^2+2ab+2ac+2bc}{a^2+b^2+c^2+2ab+2ac+2bc} \Longleftrightarrow \frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2a}\geq 1 \)
Deci inegalitatea este demonstrata.
Din CBS avem ca \( (\frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2a})[a(a+2b)+b(b+2c)+c(c+2a)]\geq (a+b+c)^2 \Longleftrightarrow \frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2a}\geq \frac{a^2+b^2+c^2+2ab+2ac+2bc}{a^2+b^2+c^2+2ab+2ac+2bc} \Longleftrightarrow \frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2a}\geq 1 \)
Deci inegalitatea este demonstrata.