Inegalitate conditionata cu ab+bc+ca=3

[Mateescu Constantin]

Fie \( a,\ b,\ c \) trei numere reale pozitive astfel incat \( ab+bc+ca=3 \).
Demonstrati ca: \( \(1+a^{2}\)\(1+b^{2}\)\(1+c^{2}\)\geq 8. \)

[Marius Mainea]

Notam \( a=\sqrt{\frac{3x^2}{xy+yz+zx}} \) si analoagele si astfel trebuie sa aratam ca

\( (3x^2+\sum xy)(3y^2+\sum xy)(2z^2+\sum xy)\ge 8(\sum xy)^3 \)

Aceasta se reduce la

\( \sum_{sym}x^3y^3+3\sum_{sym} x^4y^2+3\sum_{sym} x^4yz\ge 6\sum_{sym} x^3y^2z+\sum_{sym} x^2y^2z^2 \)

care este adevarata.

[Mateescu Constantin]

Aplicam inegalitatea lui Holder si avem:

\( (a^{2}b^{2}+a^{2}+b^{2}+1)(b^{2}+c^{2}+b^{2}c^{2}+b^{2}c^{2}+1)(a^{2}+a^{2}c^{2}+c^{2}+1)\geq \left\(\sqrt[4]{a^{2}b^{2}b^{2}a^{2}}+\sqrt[4]{a^{2}c^{2}a^{2}c^{2}}+\sqrt[4]{b^{2}b^{2}c^{2}c^{2}}+\sqrt[4]{1\cdot1\cdot1}\right\)^{4}=(ab+bc+ca+1)^{4}=64. \)

\( \Longleftrightarrow (a^{2}+1)^{2}(b^{2}+1)^{2}(c^{2}+1)^{2}\geq 64. \)

\( \Longleftrightarrow (a^{2}+1)(b^{2}+1)(c^{2}+1)\geq 8. \)

[alex2008]

Inegalitatea este echivalenta cu :
\( (1+a^2)(1+b^2)(1+c^2)\ge (ab+bc+ca-1)^2 \)

Folosim \( AM-GM \) si avem ca :

\( RHS=a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)-2(ab+bc+ca)+1\le a^2b^2+b^2c^2+c^2a^2+a^2b^2c^2+(a+b+c)^2-2(ab+bc+ca)+1=LHS \)