Find minimum value of $P=17x^2+17y^2+16xy$ Let $x,y$ be positive real numbers such that $4x^2+4y^2+17xy+5x+5y\ge 1$. Find minimum value of $$P=17x^2+17y^2+16xy$$

My idea:
I see that $x=y=\frac{\sqrt 2 -1}{5}\rightarrow P=2(3-2\sqrt 2)$
So I proved $P\ge 2(3-2 \sqrt 2)(4x^2+4y^2+17xy+5x+5y)\ge2(3-2\sqrt 2)$
Is my approachment is true?

We have
$$17\,x^{\,2}+ 17\,y^{\,2}+ 16\,xy\geqq 2(\,3- 2^{\,\frac{3}{2}}\,)$$
$$\because\,17\,x^{\,2}+ 17\,y^{\,2}+ 16\,xy- 2(\,3- 2^{\,\frac{3}{2}}\,)- (\,2- 2^{\,\frac{1}{2}}\,)(\,\underbrace{4\,x^{\,2}+ 4\,y^{\,2}+ 17\,xy+ 5\,x+ 5\,y- 1}_{\,\geqq 0}\,)\geqq 0$$
$$\because\,{\rm discriminant}[\,17\,x^{\,2}+ 17\,y^{\,2}+ 16\,xy- 2(\,3- 2^{\,\frac{3}{2}}\,)- (\,2- 2^{\,\frac{1}{2}}\,)(\,4\,x^{\,2}+ 4\,y^{\,2}+ 17\,xy+ 5\,x+ 5\,y- 1\,),\,x]= -\,\frac{126\,(\,1+ 2^{\,\frac{3}{2}}\,)(\,-\,5\,y+ 2^{\,\frac{1}{2}}- 1\,)^{\,2}}{9+ 2^{\,\frac{5}{2}}}\leqq 0$$
