JBTST III 2008, Problema 2

[Laurian Filip]

Fie a,b,c numere reale pozitive cu \( ab+bc+ca=3 \). Sa se arate ca

\( \frac{1}{1+a^2(b+c)}+\frac{1}{1+b^2(c+a)}+\frac{1}{1+c^2(a+b)} \leq \frac{1}{abc} \)

[mihai++]

\( \sum\frac{1}{1+a^2\left(b+c\right)}\leq\frac{1}{abc}\leftrightarrow \sum\frac{1}{\frac{1}{abc}+a\left(\frac{1}{b}+\frac{1}{c}\right)}\leq1\leftrightarrow\sum\frac{1}{1+a\left(\frac{1}{b}+\frac{1}{c}\right)}\leq1 \)
Am folosit \( abc\leq1 \) care iese usor din ipoteza.
\( \leftrightarrow \sum\frac{1}{1+a\left(\frac{3}{abc}-\frac{1}{a}\right)}\leq1
\leftrightarrow \sum\frac{bc}{4-bc}\leq1 \).
Am folosit ipoteza impartind prin \( abc \).
Functia \( f(x)=\frac{x}{4-x} \)este strict convexa pe \( \left(0,\infty\right) \) si aplicand majorizare ( ineg Karamata ):
\( f(bc)+f(ca)+f(ab)\leq f(ab+bc+ca)+f(0)+f(0)=\frac{ab+bc+ca}{4-ab-bc-ca}=3 \).
egalitate pt \( a=b=c=1 \).

[Marius Mainea]

\( LHS\leq\sum {\frac{1}{abc+a^2(b+c)}}=\frac{1}{3}\sum\frac{1}{a}=\frac{1}{abc} \) deoarece \( 1\geq abc \) conform \( AM\geq GM \)