Minimum of f(a,b)

Suppose that $X=a^2$ and $Y=b^3$ , then suppose that $f_{1} (X)=2X^4+X^2-2X$ and $f_{2} (Y)=2Y^2-Y-2$ . In order to get the minimum value of $f_{1} (X)+f_{2} (Y)$ using algebraic principles and concepts, we can use the vertex form rule in both equations to find our desired values. Applying the completing the square method, we have $f_{1} (X)=2X^4+X^2-2X=2X^4-X^2+2X^2-2X⟺f_1 (X)=2(X^4-1/2 X^2 )+2(X^2-X)$ ⟺ $f_{1} (X)=2(X^4-1/2 X^2+1/16)+2(X^2-X+1/4)-1/8-1/2$ ⟺ $f_{1} (X)=2(X^2-1/4)^2+2(X-1/2)^2-5/8$ and $f_{2} (Y)=2Y^2-Y-2=2(Y^2-1/2 Y)-2$ ⟺ $f_{2} (Y)=2(Y^2-1/2 Y+1/16)-2- 1/8$ ⟺ $f_{2} (Y)=2(Y-1/4)^2-17/8$ . Thus, $min_{X,Y∈R}⁡{f_{1} (X)+f_{2} (Y) }=min_{X∈R}⁡{f_{1} (X) }+min_{Y∈R}⁡{f_{2} (Y) }=-5/8-17/8=-11/4$ and the minimum value occurs when $X= \frac{1}{2}$ ; $Y=\frac{1}{4}$ $⟹$ $a=2^{-\frac{1}{2}}$ ; $b=2^{-\frac{2}{3}}$ .