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

If x,y∈R x^2 −xy+4=0 ax^2 +(b+y^2 )x+4a=0 Find the minimum of a^2 +(1/2)b^2

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

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

case 1: a=0 from (ii): (b+y^2 )x=0 ⇒x=0 ⇒impossible, otherwise 4=0! ⇒b+y^2 =0 from (i): Δ=y^2 −4×4≥0 −b−16≥0 b≤−16 a^2 +(b^2 /2)=(b^2 /2)≥((16^2 )/2)=128 case 2: a≠0 Δ=y^2 −4×4≥0 ⇒y^2 ≥16⇒y≤−4 or ≥4 ax^2 −axy+4a=0 ...(iii) (ii)−(iii): (b+y^2 )x+axy=0 x(y^2 +ay+b)=0 ⇒x=0 ⇒impossible, otberwise 4=0! ⇒y^2 +ay+b=0 Δ=a^2 −4b≥0 y_1 =((−a−(√Δ))/2), y_2 =((−a+(√Δ))/2) case 2.1:(√()) y_2 =((−a+(√Δ))/2)≤−4 −a+(√Δ)≤−8 a≥8+(√Δ) ⇒a≥8 and b≥16 ⇒a^2 +(b^2 /2)≥192 case 2.2:(√()) y_1 =((−a−(√Δ))/2)≥4 a(√())≤−8−(√Δ) ⇒a≤−8 and b≤−16 ⇒a^2 +(b^2 /2)≥192 case 2.3:(√()) y_1 =((−a−(√Δ))/2)≤−4 a≥8−(√Δ) y_2 =((−a+(√Δ))/2)≥4 a≤−(8−(√Δ)) ⇒a=0, b=−16

$${case}\:\mathrm{1}:\:{a}=\mathrm{0} \\ $$$${from}\:\left({ii}\right): \\ $$$$\left({b}+{y}^{\mathrm{2}} \right){x}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{0}\:\Rightarrow{impossible},\:{otherwise}\:\mathrm{4}=\mathrm{0}! \\ $$$$\Rightarrow{b}+{y}^{\mathrm{2}} =\mathrm{0} \\ $$$${from}\:\left({i}\right): \\ $$$$\Delta={y}^{\mathrm{2}} −\mathrm{4}×\mathrm{4}\geqslant\mathrm{0} \\ $$$$−{b}−\mathrm{16}\geqslant\mathrm{0} \\ $$$${b}\leqslant−\mathrm{16} \\ $$$${a}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} }{\mathrm{2}}=\frac{{b}^{\mathrm{2}} }{\mathrm{2}}\geqslant\frac{\mathrm{16}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{128} \\ $$$$ \\ $$$${case}\:\mathrm{2}:\:{a}\neq\mathrm{0} \\ $$$$\Delta={y}^{\mathrm{2}} −\mathrm{4}×\mathrm{4}\geqslant\mathrm{0}\:\Rightarrow{y}^{\mathrm{2}} \geqslant\mathrm{16}\Rightarrow{y}\leqslant−\mathrm{4}\:{or}\:\geqslant\mathrm{4} \\ $$$${ax}^{\mathrm{2}} −{axy}+\mathrm{4}{a}=\mathrm{0}\:\:\:...\left({iii}\right) \\ $$$$\left({ii}\right)−\left({iii}\right): \\ $$$$\left({b}+{y}^{\mathrm{2}} \right){x}+{axy}=\mathrm{0} \\ $$$${x}\left({y}^{\mathrm{2}} +{ay}+{b}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{0}\:\Rightarrow{impossible},\:{otberwise}\:\mathrm{4}=\mathrm{0}! \\ $$$$\Rightarrow{y}^{\mathrm{2}} +{ay}+{b}=\mathrm{0} \\ $$$$\Delta={a}^{\mathrm{2}} −\mathrm{4}{b}\geqslant\mathrm{0} \\ $$$${y}_{\mathrm{1}} =\frac{−{a}−\sqrt{\Delta}}{\mathrm{2}},\:{y}_{\mathrm{2}} =\frac{−{a}+\sqrt{\Delta}}{\mathrm{2}} \\ $$$${case}\:\mathrm{2}.\mathrm{1}:\sqrt{} \\ $$$${y}_{\mathrm{2}} =\frac{−{a}+\sqrt{\Delta}}{\mathrm{2}}\leqslant−\mathrm{4} \\ $$$$−{a}+\sqrt{\Delta}\leqslant−\mathrm{8} \\ $$$${a}\geqslant\mathrm{8}+\sqrt{\Delta} \\ $$$$\Rightarrow{a}\geqslant\mathrm{8}\:{and}\:{b}\geqslant\mathrm{16} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} }{\mathrm{2}}\geqslant\mathrm{192} \\ $$$${case}\:\mathrm{2}.\mathrm{2}:\sqrt{} \\ $$$${y}_{\mathrm{1}} =\frac{−{a}−\sqrt{\Delta}}{\mathrm{2}}\geqslant\mathrm{4} \\ $$$${a}\sqrt{}\leqslant−\mathrm{8}−\sqrt{\Delta} \\ $$$$\Rightarrow{a}\leqslant−\mathrm{8}\:{and}\:{b}\leqslant−\mathrm{16} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} }{\mathrm{2}}\geqslant\mathrm{192} \\ $$$${case}\:\mathrm{2}.\mathrm{3}:\sqrt{} \\ $$$${y}_{\mathrm{1}} =\frac{−{a}−\sqrt{\Delta}}{\mathrm{2}}\leqslant−\mathrm{4} \\ $$$${a}\geqslant\mathrm{8}−\sqrt{\Delta} \\ $$$${y}_{\mathrm{2}} =\frac{−{a}+\sqrt{\Delta}}{\mathrm{2}}\geqslant\mathrm{4} \\ $$$${a}\leqslant−\left(\mathrm{8}−\sqrt{\Delta}\right) \\ $$$$\Rightarrow{a}=\mathrm{0},\:{b}=−\mathrm{16} \\ $$

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

answer is 128

$${answer}\:{is}\:\mathrm{128} \\ $$

Commented by ZiYangLee last updated on 15/Aug/20

answer is ((128)/9)

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

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