7. Zadatak

Podsjetimo se rastava na faktore

$$1+a^2+a^4=1+2a^2+a^4-a^2=(1+a^2)^2-a^2=(1+a+a^2)(1-a+a^2).$$

Tada je dana nejednakost ekvivalentna sljedećem nizu nejednakosti:

$$\begin{aligned} 3(1+a^2+a^4)-(1+a+a^2)^2&\ge0\\ \Longleftrightarrow\quad 3(1+a+a^2)(1-a+a^2)-(1+a+a^2)^2&\ge0\\ \Longleftrightarrow\quad (1+a+a^2)[3(1-a+a^2)-(1+a+a^2)]&\ge0\\ \Longleftrightarrow\quad (1+a+a^2)[3-3a+3a^2-1-a-a^2]&\ge0\\ \Longleftrightarrow\quad (1+a+a^2)[2a^2-4a+2]&\ge0\\ \Longleftrightarrow\quad (1+a+a^2)2(a-1)^2&\ge0\\ \Longleftrightarrow\quad (1+a+a^2)(a-1)^2&\ge0. \end{aligned}$$

Posljednja je nejednakost uvijek točna jer je

$$1+a+a^2>0\quad\text{za sve }a\in\mathbb R$$

(jer $D<0$ ili $1+a+a^2=\left(a+\frac12\right)^2+\frac34>0$) i $(a-1)^2\ge0$ za sve $a\in\mathbb R$.

Jednakost vrijedi samo ako je $a-1=0$, odnosno za $a=1$.

Napomena:

Do ekvivalentne nejednakosti $(1+a+a^2)(a-1)^2\ge0$ moglo se doći od početne i ovako:

$$\begin{aligned} 3(1+a^2+a^4)-(1+a+a^2)^2&\ge0\\ \Longleftrightarrow\quad 3+3a^2+3a^4-1-a^2-a^4-2a-2a^2-2a^3&\ge0\\ \Longleftrightarrow\quad 2a^4-2a^3-2a+2&\ge0\\ \Longleftrightarrow\quad a^4-a^3-a+1&\ge0\\ \Longleftrightarrow\quad a^3(a-1)-(a-1)&\ge0\\ \Longleftrightarrow\quad (a-1)(a^3-1)&\ge0\\ \Longleftrightarrow\quad (a-1)^2(a^2+a+1)&\ge0. \end{aligned}$$