Inegalitate conditionata 5

[Marius Mainea]

Fie a,b,c>0 astfel incat \( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1 \). Demonstrati ca:

\( \frac{a^3}{(b+c)(a+1)}+\frac{b^3}{(c+a)(b+1)}+\frac{c^3}{(a+b)(c+1)}\ge \frac{27}{8}. \)

[maxim bogdan]

Din Inegalitatea Cauchy-Buniakowski-Schwarz avem: \( \sum_{cyc}\frac{a^3}{(b+c)(a+1)}=\sum_{cyc}\frac{a^2}{b+c+\frac{b+c}{a}}\geq \frac{(a+b+c)^2}{2(a+b+c)+\sum_{cyc}\frac{b+c}{a}}=\frac{(a+b+c)^2}{2(a+b+c)+(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-3)}=\frac{(a+b+c)^2}{3(a+b+c)-3}\geq\frac{27}{8} \)

Notam \( x=a+b+c. \) Evident din Inegalitatea mediilor(AM-HM) obtinem ca: \( x\geq 9 \)

Inegalitatea este echivalenta cu: \( 8x^2-81x+81\geq 0. \) Fie \( f:[9,\infty)\to\mathbb{R} \), cu \( f(x)=8x^2-81x+81 \)

Avem: \( \Delta=63^2. \) Deci \( x_{1}=\frac{9}{8}<9=x_{2} \)

Deci \( f \) ia valori pozitive pe intervalul \( [9,\infty). \)