Stiind ca \( abc=1 \), \( a,b,c\in\mathbb{R}_+^* \), aratati ca
\( \sum_{cyc}{\frac{1}{a^2+2b^2+3}\le\frac{1}{2} \).
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Solutie!
Avem: \( \sum_{cyc}\frac{1}{a^2+2b^2+3}=\sum_{cyc}\frac{1}{(a^2+b^2)+(b^2+1)+2}\leq \sum_{cyc}\frac{1}{2ab+2b+2} \) (Inegalitatea mediilor).
Cum \( abc=1\Rightarrow(\exist) \) \( x,y,z\in\mathb{R}_{+}^* \) astfel incat \( a=\frac{x}{y}; b=\frac{y}{z}; c=\frac{z}{x}. \)
De aici va rezulta ca: \( \frac{1}{2}(\sum_{cyc}\frac{1}{ab+b+1})=\frac{1}{2}(\sum_{cyc}\frac{z}{x+y+z})=\frac{1}{2}. \)
Deci inegalitatea noastra este demonstrata!
Cum \( abc=1\Rightarrow(\exist) \) \( x,y,z\in\mathb{R}_{+}^* \) astfel incat \( a=\frac{x}{y}; b=\frac{y}{z}; c=\frac{z}{x}. \)
De aici va rezulta ca: \( \frac{1}{2}(\sum_{cyc}\frac{1}{ab+b+1})=\frac{1}{2}(\sum_{cyc}\frac{z}{x+y+z})=\frac{1}{2}. \)
Deci inegalitatea noastra este demonstrata!
Feuerbach