Find minimum value of $P=17x^2+17y^2+16xy$

Question

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

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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?

Answer 1

Yes, your answer is true.

The solution can be the following.

By AM-GM $$1\leq4(x+y)^2+5(x+y)+9xy\leq4(x+y)^2+5(x+y)+\frac{9}{4}(x+y)^2,$$ which gives $$x+y\geq\frac{2(\sqrt2-1)}{5}.$$ Thus, by C-S we obtain: $$17(x^2+y^2)+16xy=8(x+y)^2+\frac{9}{2}(1^2+1^2)(x^2+y^2)\geq8(x+y)^2+\frac{9}{2}(x+y)^2=$$ $$=\frac{25}{2}(x+y)^2\geq\frac{25}{2}\cdot\frac{4(3-2\sqrt2)}{25}=2(3-2\sqrt2).$$ The equality occurs for $x=y=\frac{\sqrt2-1}{5},$ which says that we got a minimal value.

Answer 2

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