ΛΥΣΗ

α) Είναι:

$${\rm K}-\Lambda =2{\alpha }^2+{\beta }^2+9-2\alpha (3-\beta )$$ $$=2{\alpha }^2+{\beta }^2+9-(6\alpha -2\alpha \beta )$$ $$={\alpha }^2+{\alpha }^2+{\beta }^2+9-6\alpha +2\alpha \beta$$ $$=({\alpha }^2+2\alpha \beta +{\beta }^2)+({\alpha }^2-6\alpha +9)$$

β) Ισοδύναμα και διαδοχικά ισχύει ότι:

$${\rm K}\ge \Lambda$$ $$\iff {\rm K}-\Lambda \ge 0$$ $$\iff ({\alpha }^2+2\alpha \beta +{\beta }^2)+({\alpha }^2-6\alpha +9)\ge 0$$ $$\iff (\alpha +\beta {)}^2+(\alpha -3{)}^2\ge 0$$

το οποίο ισχύει για κάθε τιμή των $\alpha$, $\beta$.

γ) Η ισότητα ισχύει αν και μόνο αν:

$${\rm K}=\Lambda$$ $$\iff {\rm K}-\Lambda =0$$ $$\iff (\alpha +\beta {)}^2+(\alpha -3{)}^2=0$$ $$\iff \{\begin{array}{l}(\alpha +\beta {)}^2=0\\ (\alpha -3{)}^2=0\end{array}$$ $$\iff \{\begin{array}{l}\alpha +\beta =0\\ \alpha -3=0\end{array}$$ $$\iff \{\begin{array}{l}\alpha =-\beta \\ \alpha =3\end{array}$$ $$\iff \{\begin{array}{l}\alpha =3\\ \beta =-3\end{array}$$