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If-xy-ax-2-2y-2-is-always-true-for-any-1-x-2-2-y-3-Then-find-the-range-of-a-

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Question Number 184618 by CrispyXYZ last updated on 09/Jan/23
If xy≤ax^2 +2y^2 is always true for any 1≤x≤2, 2≤y≤3 Then find the range of a.
$$\mathrm{If}\:{xy}\leqslant{ax}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{always}\:\mathrm{true}\:\mathrm{for}\:\mathrm{any}\:\mathrm{1}\leqslant{x}\leqslant\mathrm{2},\:\mathrm{2}\leqslant{y}\leqslant\mathrm{3} \\ $$$$\mathrm{Then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{a}. \\ $$
Answered by mr W last updated on 09/Jan/23
let t=(y/x) 2t^2 −t+a≥0 1≤t≤3 case (A) Δ=1−8a≤0 ⇒a≥(1/8) or Δ=1−8a≥0 ⇒a≤(1/8) and case (C) case (B) ((1−(√(1−8a)))/4)≥3 or ((1+(√(1−8a)))/4)≤1 −(√(1−8a))≥^(never) 11 or a≥−1 summary: a≥(1/8) or −1≤a≤(1/8), i.e. a≥−1
$${let}\:{t}=\frac{{y}}{{x}} \\ $$$$\mathrm{2}{t}^{\mathrm{2}} −{t}+{a}\geqslant\mathrm{0} \\ $$$$\mathrm{1}\leqslant{t}\leqslant\mathrm{3} \\ $$$${case}\:\left({A}\right) \\ $$$$\Delta=\mathrm{1}−\mathrm{8}{a}\leqslant\mathrm{0}\:\Rightarrow{a}\geqslant\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${or} \\ $$$$\Delta=\mathrm{1}−\mathrm{8}{a}\geqslant\mathrm{0}\:\Rightarrow{a}\leqslant\frac{\mathrm{1}}{\mathrm{8}}\:{and} \\ $$$${case}\:\left({C}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{case}\:\left({B}\right) \\ $$$$\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{8}{a}}}{\mathrm{4}}\geqslant\mathrm{3}\:{or}\:\frac{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{8}{a}}}{\mathrm{4}}\leqslant\mathrm{1} \\ $$$$−\sqrt{\mathrm{1}−\mathrm{8}{a}}\overset{{never}} {\geqslant}\mathrm{11}\:{or}\:{a}\geqslant−\mathrm{1} \\ $$$$ \\ $$$${summary}:\: \\ $$$${a}\geqslant\frac{\mathrm{1}}{\mathrm{8}}\:{or}\:−\mathrm{1}\leqslant{a}\leqslant\frac{\mathrm{1}}{\mathrm{8}},\:{i}.{e}.\:{a}\geqslant−\mathrm{1} \\ $$
Commented by SEKRET last updated on 09/Jan/23
−15≤a≤(1/8)
$$\:−\mathrm{15}\leqslant\boldsymbol{\mathrm{a}}\leqslant\frac{\mathrm{1}}{\mathrm{8}} \\ $$
Commented by mr W last updated on 10/Jan/23
Commented by mr W last updated on 09/Jan/23
if −15≤a≤(1/8) is right, then a=1 is wrong. but 2t^2 −t+1 is always >0 for 1≤t≤3. that means a=1 is right.
$${if}\:−\mathrm{15}\leqslant{a}\leqslant\frac{\mathrm{1}}{\mathrm{8}}\:{is}\:{right},\:{then}\:{a}=\mathrm{1}\:{is} \\ $$$${wrong}.\:{but}\:\mathrm{2}{t}^{\mathrm{2}} −{t}+\mathrm{1}\:{is}\:{always}\:>\mathrm{0} \\ $$$${for}\:\mathrm{1}\leqslant{t}\leqslant\mathrm{3}.\:{that}\:{means}\:{a}=\mathrm{1}\:{is}\:{right}. \\ $$
Commented by mr W last updated on 09/Jan/23
if −15≤a≤(1/8) is right, then a=−15 is right, but 2t^2 −t−15 is <0 for 1≤t≤3, that means a=−15 is wrong.
$${if}\:−\mathrm{15}\leqslant{a}\leqslant\frac{\mathrm{1}}{\mathrm{8}}\:{is}\:{right},\:{then}\:{a}=−\mathrm{15}\:{is} \\ $$$${right},\:{but}\:\mathrm{2}{t}^{\mathrm{2}} −{t}−\mathrm{15}\:{is}\:<\mathrm{0}\:{for}\:\mathrm{1}\leqslant{t}\leqslant\mathrm{3}, \\ $$$${that}\:{means}\:{a}=−\mathrm{15}\:{is}\:{wrong}. \\ $$

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