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Fie numerele reale strict pozitive \( a,\ b,\ c \) astfel incat \( a^{2}+b^{2}+c^{2}=3 \). Sa se arate ca \( \displaystyle \frac{a^{3}}{c^{2}+ab}+\frac{b^{3}}{a^{2}+bc}+\frac{c^{3}}{b^{2}+ca}\ge\frac{3}{2} \).

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\( \sum_{cyc}\frac{a^3}{c^2+ab}=\sum_{cyc}\frac{a^4}{ac^2+a^2b}\ge\frac{(a^2+b^2+c^2)^2}{\sum_{sym}a^2b}=\frac{9}{\sum_{sym}a^2b} \)
Deci ramane de demonstrat \( \sum_{sym}a^2b\le 6=2(a^2+b^2+c^2)\Longleftrightarrow \sum_{cyc}a(b^2+c^2)\le 2(a^2+b^2+c^2)\Longleftrightarrow \sum_{cyc}a(3-a^2)\le 2(a^2+b^2+c^2)\Longrightarrow a(a-1)(a+3)+b(b-1)(b+3)+c(c-1)(c+3)\ge 0 \)
Functia \( f(x)=x(x-1)(x+3) \) este convexa pe \( \(0,+\infty\) \)
Aplicand inegalitatea lui Jensen \( f(a)+f(b)+f(c)\ge 3f(\frac{a+b+c}{3})=(a+b+c)(a+b+c-1)(a+b+c+3)\ge 0; \) deoarece cel putin unul din numerele \( a,b,c\ge 1 \) si problema este rezolvata.

DrAGos Calinescu wrote:\( \sum_{cyc}\frac{a^3}{c^2+ab}=\sum_{cyc}\frac{a^4}{ac^2+a^2b}\ge\frac{(a^2+b^2+c^2)^2}{\sum_{sym}a^2b} \)

Nu prea e simetrica suma de la numitor

Da acum am vazut.
Revenind acolo
\( LHS\ge\frac{9}{2(a^2b+b^2c+c^2a)} \)
Ramane de demonstrat \( a^2b+b^2c+c^2a\le 3 \)
Dar \( a^2b+b^2c+c^2a\le\sqrt{(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2)}=\sqrt{3(a^2b^2+b^2c^2+c^2a^2)} (1) \)
\( a^2b^2+b^2c^2+c^2a^2\le\frac{(a^2+b^2+c^2)^2}{3}=3 \)
Acum revenind in \( (1) \) s-a terminat.
Sper ca de data asta sa nu mai fi gresit