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
 A: 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:

$$9=7a^2-9ab+7b^2 = \frac{23}{2}(a^2+b^2) - \frac{9}{2}(a+b)^2 \le \frac{23}{2}(a^2+b^2)$$
A: 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.
$$\color{red}{\frac{18}{23}\le a^2+b^2\le\frac{18}{5}}$$
A: Though many elegant answers have already been given so I'm just adding this to add another perspective (which can also be interpreted from some of the solutions above). 
There is a geometrical way to look at this problem. The given curve is an ellipse which is symmetric about the line $y=x$ (and $y=-x$ as well). We want to optimize $x^2+y^2$. Say $x^2+y^2=c$, then we are looking for circles which touch the given ellipse at its extremities.  So we need just the lengths of the major and minor axes of the ellipse. See figure below.

A: 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. 
