Finding range of expression $f(x,y)=x^2+y^2$ 
Finding range of $f(x,y)=x^2+y^2$ subjected to
the condition $2x^2+6xy+5y^2=1$ without Calculus

Try: Let $k=x^2+y^2,$ Then $\displaystyle k=\frac{x^2+y^2}{2x^2+6xy+5y^2}$
Now put $\displaystyle \frac{y}{x}=t(x\neq 0)$ and $t\in \mathbb{R}$
$$k=\frac{1+t^2}{2+6t+5t^2}\Rightarrow (5k-1)t^2+6kt+(2k-1)=0$$
For real $t$ its Discriminant $\geq 0$
$$36k^2-4(5k-1)(2k-1)\geq 0$$
$$k^2-7k+1\leq 0$$
$$\Rightarrow k\in \bigg[\frac{7-3\sqrt{5}}{2}\;\;,\frac{7+3\sqrt{5}}{2}\bigg]$$
I have a question . can i solve it using Inequality like
$(a+b)^2\geq 0$
or $(a-b)^2\geq 0$ , explain me thanks

Yes, of course. We can make it. 
For example, we need to prove that
$$x^2+y^2\geq\frac{7-3\sqrt5}{2}(2x^2+6xy+5y^2)$$ or
$$(2\sqrt5-4)x^2-2(7-3\sqrt5)xy+(5\sqrt5-11)y^2\geq0$$ or
$$2x^2-2(7-3\sqrt5)(2+\sqrt5)xy+(5\sqrt5-1)(2+\sqrt5)y^2\geq0$$ or
$$2x^2-2(-1+\sqrt5)xy+(3-\sqrt5)y^2\geq0$$ or
$$4x^2-4(-1+\sqrt5)xy+(6-2\sqrt5)y^2\geq0$$ or
$$(2x-(\sqrt5-1)y)^2\geq0.$$
The second inequality we can prove by the similar way.
