Sa se arate ca x² +4x+5 ≥ 1 / (x²+2x+2) , oricare ar fi x∈R
As dori raspuns complet , Multumesc!


Răspuns :

[tex]x^2+2x+2=(x^2+2x+1)+1=(x+1)^2+1 \geq 1. \\ \\ Deci~ \frac{1}{x^2+2x+2} \leq \frac{1}{1}=1..................(1) \\ \\ x^2+4x+5=(x^2+4x+4)+1 =(x+2)^2+1 \geq 1..................(2) \\ \\ Din~(1)~si~(2)~rezulta~ca~x^2+4x+5 \geq \frac{1}{x^2+2x+2}. \\ \\ \underline{Observatie}:~egalitatea~nu~poate~avea~loc. [/tex]