<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] If $a = 0$ or $b = 0$, then $a^2+b^2 = \dfrac{9}{7}$, and this is one of the many values in the range of $a^2+b^2$. So assume $ab \neq 0$, then put $x = \dfrac{a}{b} > 0$ ( we can assume $a, b > 0$ ), and rewrite the equation by dividing both sides by $ab$ to have: $\dfrac{9}{ab} = 7\left(x+\dfrac{1}{x}\right)-9 = 7u-9, u = x+\dfrac{1}{x}\implies a^2+b^2 = ab\left(\dfrac{a}{b}+\dfrac{b}{a}\right) = ab\left(x+\dfrac{1}{x}\right) = abu = \dfrac{9u}{7u-9}$. Thus consider $f(u) = \dfrac{9u}{7u-9}, u \ge 2\implies f'(u) = \dfrac{9(7u-9)-7(9u)}{(7u-9)^2}= \dfrac{-81}{(7u-9)^2} < 0\implies f(u) \le f(2)= \dfrac{18}{5}$. Thus $(a^2+b^2)_{\text{max}} = \dfrac{18}{5} $. Note that the OP has made an edit to his answer, and I thought his first answer to the minimum value was incorrect, but due to trusting his own answer, I didn't validate it but it was confirmed by the 3rd answer to this question that it was correct. 
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