<s>[INST] Minimum value of expression having $2$ variables 
Minimum value of $$\bigg(x-4-\sqrt{4-y^2}\bigg)^2+\bigg(4\sqrt{x}-y\bigg)^2$$ for real $x\geq 0,y\in[-2,2]$

Try: Using Partial derivative 
$$f(x,y) = \bigg(x-4-\sqrt{4-y^2}\bigg)^2+\bigg(4\sqrt{x}-y\bigg)^2$$
$$\frac{df(x,y)}{dx}=2\bigg(x-4-\sqrt{4-y^2}\bigg)+\frac{4}{\sqrt{x}}\bigg(4\sqrt{x}-y\bigg)$$
$$\frac{df(x,y)}{dy}=-2y\frac{\bigg(x-4-\sqrt{4-y^2}\bigg)}{\sqrt{4-y^2}}-2\bigg(4\sqrt{x}-y\bigg)$$
Put $\displaystyle \frac{df}{dx} =0$ and $\displaystyle \frac{df}{dy}=0$
But these $2$ equation is tedious work
Could some help me how to solve it , Thanks in advance
 [/INST] For $0\leq x\leq 4$ we have 
$$f(x,y)-f(x,2)=2\left(4\sqrt{x}(2-y)+(4-x)\sqrt{4-y^2}\right)\geq 0 $$
$$f(x,y)\geq  f(x,2)\geq  \underset{0\leq x\leq 4}{\min }f(x,2)= \underset{0\leq x\leq 4}{\min }\left(20-16 \sqrt{x}+8 x+x^2\right)$$
For $x\geq 4$, use the Cauchy–Schwarz inequality, to show $f(x,y) \geq 36$ with $f(4,2)=36$ as follows
$$\left(x-4,4\sqrt{x}\right).\left(\sqrt{4-y^2},y\right)\leq  2(x+4)$$
$$f(x,y)-36=-16+8 x+x^2-2\left((x-4) \sqrt{4-y^2}+4 \sqrt{x} y\right)$$
$$\geq  -16+8 x+x^2-4(x+4)=(x-4)(x+8)\geq 0$$
EDIT:
$\underset{0\leq x\leq 4}{\min }\left(20-16 \sqrt{x}+8 x+x^2\right)$ occurs at the zero of the cubic polynomial $x^3+8x^2+16x-16$ with $x \approx 0.7186$ and the minimum value $\approx 12.7019$
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