Fie \( a,b,c\in \left[\frac{1}{\sqrt{2}},\sqrt{2}\right] \) . Sa se demonstreze ca :
\(
\sum_{cyc}\frac{3}{a+2b}\ge \sum_{cyc}\frac{2}{a+b} \)
Inegalitate intr-un interval
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Inegalitate intr-un interval
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Mateescu Constantin
- Newton
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Scriem inegalitatea sub forma:
\( \sum\left\(\frac{3}{a+2b}-\frac{2}{a+b}+\frac{1}{6a}-\frac{1}{6b}\right\)\ge 0 \)
\( \Longleftrightarrow \sum\frac{(a-b)^2(2b-a)}{6ab(a+2b)(a+b)}\ge 0 \)
Dar \( 2b-a\ge \frac{2}{\sqrt{2}}-\sqrt{2}=0 \) si astfel inegalitatea e demonstrata. Egalitatea are loc pentru \( a=b=c \).
\( \sum\left\(\frac{3}{a+2b}-\frac{2}{a+b}+\frac{1}{6a}-\frac{1}{6b}\right\)\ge 0 \)
\( \Longleftrightarrow \sum\frac{(a-b)^2(2b-a)}{6ab(a+2b)(a+b)}\ge 0 \)
Dar \( 2b-a\ge \frac{2}{\sqrt{2}}-\sqrt{2}=0 \) si astfel inegalitatea e demonstrata. Egalitatea are loc pentru \( a=b=c \).