<s>[INST] find range of $a^2+b^2$ without trigonometric substution given $7a^2-9ab+7b^2=9$ and $a,b$ are real no. then find range of $a^2+b^2$ without trigonometric substution
from $7a^2-9ab+7b^2=9$
$\displaystyle ab = \frac{7(a^2+b^2)-9}{9}$
put into inequality $\displaystyle a^2+b^2 \geq 2ab$
$\displaystyle a^2+b^2 \geq \frac{14(a^2+b^2)-18}{9}$
$\displaystyle a^2+b^2 \leq \frac{18}{5}$
i wan,t be able to find minimum,could some help me with this
 [/INST] For the maximum, note that:
$$9=7a^2-9ab+7b^2 = \frac{5}{2}(a^2+b^2) + \frac{9}{2}(a-b)^2 \ge \frac{5}{2}(a^2+b^2)$$

[ EDIT ]  For the minimum, note that:

$$9=7a^2-9ab+7b^2 = \frac{23}{2}(a^2+b^2) - \frac{9}{2}(a+b)^2 \le \frac{23}{2}(a^2+b^2)$$
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