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Question Number 108171 by ZiYangLee last updated on 15/Aug/20

$$\mathrm{If}x,y\in \mathbb{R}$$$$x^2-xy+4=0$$$${ax}^2+(b+y^2)x+4a=0$$$$\mathrm{Find}\mathrm{the}\mathrm{minimum}\mathrm{of}{a}^2+\frac{1}{2}{b}^2$$

Answered by mr W last updated on 15/Aug/20

$$case1:a=0$$$$from\left(ii\right):$$$$(b+y^2)x=0$$$$\Rightarrow x=0\Rightarrow impossible,otherwise4=0!$$$$\Rightarrow b+y^2=0$$$$from\left(i\right):$$$$\mathrm{\Delta }=y^2-4\times 4⩾0$$$$-b-16⩾0$$$$b⩽-16$$$$a^2+\frac{b^2}{2}=\frac{b^2}{2}⩾\frac{{16}^2}{2}=128$$$$$$$$case2:a\ne 0$$$$\mathrm{\Delta }=y^2-4\times 4⩾0\Rightarrow y^2⩾16\Rightarrow y⩽-4or⩾4$$$${ax}^2-axy+4a=0...\left(iii\right)$$$$\left(ii\right)-\left(iii\right):$$$$(b+y^2)x+axy=0$$$$x(y^2+ay+b)=0$$$$\Rightarrow x=0\Rightarrow impossible,otberwise4=0!$$$$\Rightarrow y^2+ay+b=0$$$$\mathrm{\Delta }=a^2-4b⩾0$$$$y_1=\frac{-a-\sqrt{\mathrm{\Delta }}}{2},y_2=\frac{-a+\sqrt{\mathrm{\Delta }}}{2}$$$$case2.1:\sqrt{}$$$$y_2=\frac{-a+\sqrt{\mathrm{\Delta }}}{2}⩽-4$$$$-a+\sqrt{\mathrm{\Delta }}⩽-8$$$$a⩾8+\sqrt{\mathrm{\Delta }}$$$$\Rightarrow a⩾8andb⩾16$$$$\Rightarrow a^2+\frac{b^2}{2}⩾192$$$$case2.2:\sqrt{}$$$$y_1=\frac{-a-\sqrt{\mathrm{\Delta }}}{2}⩾4$$$$a\sqrt{}⩽-8-\sqrt{\mathrm{\Delta }}$$$$\Rightarrow a⩽-8andb⩽-16$$$$\Rightarrow a^2+\frac{b^2}{2}⩾192$$$$case2.3:\sqrt{}$$$$y_1=\frac{-a-\sqrt{\mathrm{\Delta }}}{2}⩽-4$$$$a⩾8-\sqrt{\mathrm{\Delta }}$$$$y_2=\frac{-a+\sqrt{\mathrm{\Delta }}}{2}⩾4$$$$a⩽-(8-\sqrt{\mathrm{\Delta }})$$$$\Rightarrow a=0,b=-16$$

Commented by mr W last updated on 15/Aug/20

$$answeris128$$

Commented by ZiYangLee last updated on 15/Aug/20

$$\mathrm{answer}\mathrm{is}\frac{128}{9}$$

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