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Daca \( a,b,c \) sunt numere reale pozitive cu \( abc=1 \) , aratati ca :

\( \sum_{cyc}\frac{a}{2+bc}\ge 1 \)

Manuela Prajea

Pentru ca \( abc=1 \) putem face substitutiile \( a=\frac{x}{y} \) , \( b=\frac{y}{z} \) si \( c=\frac{z}{x} \). Inegalitatea din enunt devine \( \sum_{cyc}\frac{x^2}{2xy+y^2}\ge1 \). Din Cauchy (lema) obtinem \( \sum_{cyc}\frac{x^2}{2xy+y^2}\ge\frac{(x+y+z)^2}{x^2+y^2+z^2+2xy+2yz+2zx}=1 \) si de aici concluzia.

alex2008 wrote:Daca \( a,b,c \) sunt numere reale pozitive cu \( abc=1 \) , aratati ca \( \sum\frac{a}{2+bc}\ge 1 . \)

Dem. Notam \( s=a+b+c \) . Se observa ca \( s\ge 3 \) si \( s^2\ge 2s+3 \) . Deci \( \sum\frac{a}{2+bc}= \) \( \sum\frac {a^2}{2a+1}\stackrel{CBS}{\ge\ }\frac {(a+b+c)^2}{2(a+b+c)+3}= \) \( \frac {s^2}{2s+3}\ge 1 \) .

Virgil Nicula wrote: O usoara extindere.Daca \( \lambda\in [0,3] \) si \( a,b,c \) sunt numere reale pozitive cu \( abc=1 \) , aratati ca \( \sum\frac{a}{(3-\lambda ) +\lambda bc}\ge 1 \) .

Aplicatie. Daca \( \lambda\in [0,3] \) si \( a,b,c \) sunt numere reale pozitive cu \( abc=1 \) , aratati ca \( \sum\frac{a+1}{\left[(3-\lambda )+\lambda bc\right]\cdot\left[\lambda +(3-\lambda ) bc\right]}\ge \frac 23 . \)

Sau la nivelul clasei a 11-a cum \( abc=1=>bc=\frac{1}{a} \)si analoagele .Se verifica faptul ca functia \( f:\(0,\infty\)\rightarrow\mathbb{R},f(x)=\frac{x}{2+\frac{1}{x}} \)este crescatoare si convexa ,adica cu Jensen avem :\( f(a)+f(b)+f(c)\ge 3f(\frac{a+b+c}{3})\ge 3f((abc)^{\frac{1}{3}})=3f(1)=3\frac{1}{3}=1 \).

O demonstratie prin "spargere" ar fi cam asa :
\( \frac{a}{2+bc}=\frac{a^2}{2a+1}= \frac{(3a)^2}{9(2a+1)} \geq \frac{6a-2a-1}{9}=\frac{4a-1}{9} \) si analoagele care insumate du la \( \sum_{cyc}\frac{a}{2+bc}\ge \frac{4(a+b+c)-3}{9} \) si cum \( a+b+c \geq3 \) se obtine concluzia.

Virgil Nicula wrote: Aplicatie. Daca \( \lambda\in [0,3] \) si \( a,b,c \) sunt numere reale pozitive cu \( abc=1 \) , aratati ca \( \sum\frac{a+1}{\left[(3-\lambda )+\lambda bc\right]\cdot\left[\lambda +(3-\lambda ) bc\right]}\ge \frac 23 . \)

\( LHS\ge \sum\frac{a+1}{(\frac{3+3bc}{2})^2}=\frac{4}{9}\sum\frac{a^2}{a+1}\ge RHS \)