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

It seems that you've got $\frac{18}{23}$ as the minimum value incorrectly though the value itself is correct. (you have already corrected the error.)
Let $k=a^2+b^2$.
Then, we get
$$ab=\frac{7k-9}{9}\tag1$$
Also, from
$$k=(a+b)^2-2ab=(a+b)^2-2\times\frac{7k-9}{9}$$
we get
$$(a+b)^2=\frac{23k-18}{9}\tag2$$
Therefore, we have to have $$\frac{23k-18}{9}\ge 0\tag3$$
Under $(3)$, $a,b$ are the solutions of $$t^2\mp\sqrt{\frac{23k-18}{9}}\ t+\frac{7k-9}{9}=0$$
Since $a,b$ are real, we have to have
$$\left(\mp\sqrt{\frac{23k-18}{9}}\right)^2-4\cdot 1\cdot \frac{7k-9}{9}\ge 0\tag4$$
In order for real $a,b$ satisfying $(1)$ and $(2)$ to exist, it is necessary and sufficient that $(3)$ and $(4)$ hold, i.e.
$$\color{red}{\frac{18}{23}\le a^2+b^2\le\frac{18}{5}}$$
