Fie \( a\in \mathb{R}^*, b\in \mathb{R} \). Sa se demonstreze ca:
\( a^2+b^2+\frac{1}{a^2}+\frac{b}{a}\geq \sqrt{3} \).
Concursul Matefbc editia a 3-a problema 2
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\( \left(2a^2-\sqrt 3\right)^2+\left(2ab+1\right)^2\ge 0 \) si apoi se imparte la \( 4a^2. \)
Altfel : Notam \( c=\frac ba \) , adica \( b=ca \) . Inegalitatea propusa devine \( a^2+a^2c^2+\frac {1}{a^2}+c\ge\sqrt 3 \ \Longleftrightarrow\ a^2(1+c^2)+\frac {1}{a^2}\ge \sqrt 3-c \) .
Insa \( a^2\left(1+c^2\right)+\frac {1}{a^2}\ge 2\sqrt {1+c^2} \) (inegalitatea mediilor). Ramane de aratat ca \( (*)\ 2\sqrt {1+c^2}\ge \sqrt 3-c \) . Daca \( c\ge \sqrt 3 \) , atunci e O.K.
Fie \( c\ <\ \sqrt 3 \) . In acest caz inegalitatea (*) devine \( 4(c^2+1)\ge c^2-2c\sqrt 3+3 \) , adica \( 3c^2+2c\sqrt 3+1\ge 0 \Longleftrightarrow\ (c\sqrt 3+1)^2\ge 0 \) .
Altfel : Notam \( c=\frac ba \) , adica \( b=ca \) . Inegalitatea propusa devine \( a^2+a^2c^2+\frac {1}{a^2}+c\ge\sqrt 3 \ \Longleftrightarrow\ a^2(1+c^2)+\frac {1}{a^2}\ge \sqrt 3-c \) .
Insa \( a^2\left(1+c^2\right)+\frac {1}{a^2}\ge 2\sqrt {1+c^2} \) (inegalitatea mediilor). Ramane de aratat ca \( (*)\ 2\sqrt {1+c^2}\ge \sqrt 3-c \) . Daca \( c\ge \sqrt 3 \) , atunci e O.K.
Fie \( c\ <\ \sqrt 3 \) . In acest caz inegalitatea (*) devine \( 4(c^2+1)\ge c^2-2c\sqrt 3+3 \) , adica \( 3c^2+2c\sqrt 3+1\ge 0 \Longleftrightarrow\ (c\sqrt 3+1)^2\ge 0 \) .