Q108815-19-8-20-unanswer-by-1x-x-Given-f-x-1-1-x-1-1-a-ax-ax-8-x-a-R-x-a-gt-0-Prove-that-1-lt-f-x-lt-2-Solution-Put-x-tan-2-A-a-tan-2-B-A-B-0-pi-2-f-x-cosA-c

Question Number 108914 by 1549442205PVT last updated on 21/Aug/20

Q108815(19/8/20)(unanswer)by 1x.x Given f(x)=(1/( (√(1+x))))+(1/( (√(1+a))))+((√(ax))/( (√(ax+8)))) x,a∈R;x,a>0.Prove that 1<f(x)<2 Solution:Put x=tan^2 A,a=tan^2 B(A,B∈[0,(π/2)) f(x)=cosA+cosB+((tanAtanB)/( (√(tan^2 Atan^2 B+8)))) =cosA+cosB+((sinAsinB)/( (√(8cos^2 Acos^2 B+sin^2 Asin^2 B)))) Put cosA=z,cosB=y(z,y∈(0,1])we have f=z+y+((√((1−z^2 )(1−y^2 )))/( (√(8z^2 y^2 +(1−z^2 )(1−y^2 ))))) =z+y+((√((1−z^2 )(1−y^2 )))/( (√(9z^2 y^2 +1−(z^2 +y^2 ))))) i)First we prove f(x)>1 ⇔z+y+((√((1−z^2 )(1− y^2 )))/( (√(9z^2 y^2 +1−(z^2 +y^2 )))))>1(1) If z+y≥1 then the inequality(1) is true.Consider z+y<1.Put m=1−(z+y)⇔z+y=1−m(0<m≤1) z^2 +y^2 =(z+y)^2 −2zy=(1−m)^2 −2zy (1)⇔(((1−z^2 )(1−y^2 ))/( 9z^2 y^2 +1−(z^2 +y^2 )))>[1−(z+y)]^2 ⇔1+z^2 y^2 −(z^2 +y^2 )>[9z^2 y^2 +1−(z^2 +y^2 )]m^2 1+z^2 y^2 −[1−2m+m^2 −2zy]>[9z^2 y^2 +1−(1−2m+m^2 −2zy)]m^2 ⇔z^2 y^2 +2zy+2m−m^2 >(9z^2 y^2 +2zy+2m−m^2 )m^2 ⇔m^4 −m^2 (9z^2 y^2 +2zy+2m)+(z^2 y^2 +2zy+2m−m^2 )>0 We look at LHS as a quadratic polynomial with respect to “im^2 ” defined on the interval(0;1) and we denote by P(m).By the theorem above the sign of quadratic poly.P(m)>0⇔ Δ_P =(9z^2 y^2 +2zy+2m)^2 −4(z^2 y^2 +2zy+2m−m^2 )<0 ⇔81z^4 y^4 +4z^2 y^2 +4m^2 +36z^3 y^3 +36mz^2 y^2 +8mzy−4(z^2 y^2 +2zy+2m−m^2 )<0 ⇔81z^4 y^4 +36z^3 y^3 +36mz^2 y^2 +8(m−1)zy+8m^2 −8m<0 ⇔8m^2 +(36z^2 y^2 +8zy−8)m+81z^4 y^4 +36z^3 y^3 −8zy<0(3) We look at LHS (3) like as a quadratic polynomial w.r.t “m” and denote by Q(m) We has Q(0)=81(zy)^4 +36(zy)^3 −8zy ≤81t/64+36t/16−8t=225t/64−8t<0 (due to 0< t=zy≤1/4 ) Q(1)=1+36(zy)^2 +8zy−8+81(zy)^4 +36(zy)^3 −8zy =−7+36(zy)^2 +81(zy)^4 +36(zy)^3 ≤ −7+36/16+81/256+36/64<0 (due to zy≤1/4) By the convert theorem above the sign of the quadratic polynomial we infer Q(m)>0 ∀m∈(0;1)which means P(m) has Δ_P <0 ∀m∈(0;1)⇒the inequality (1) proved ii)Now we prove that f(x)<2 ⇔z+y+((√((1−z^2 )(1−y^2 )))/( (√(9z^2 y^2 +1−(z^2 +y^2 )))))<2(4) ⇔(((1−z^2 )(1−y^2 ))/(9z^2 y^2 +1−(z^2 +y^2 )))<[2−(z+y)]^2 ⇔1+z^2 y^2 −(z^2 +y^2 )<[9z^2 y^2 +1−(z^2 +y^2 )](1+m)^2 (note (1−m=z+y like as above we have −1≤m=1−(z+y)<1 as 0<z+y≤2(∗)) ⇔z^2 y^2 +2zy+2m−m^2 <(9z^2 y^2 +2zy+2m−m^2 )(1+2m+m^2 ) ⇔z^2 y^2 +2zy+2m−m^2 <(9z^2 y^2 +2zy+2m−m^2 )+(9z^2 y^2 +2zy)(2m+m^2 )−m^4 +4m^2 >0 ⇔−m^4 +4m^2 +(9z^2 y^2 +2zy)(2m+m^2 )+8z^2 y^2 >0 ⇔(9m^2 +18m+8)(zy)^2 +2(m^2 +2m)zy−m^4 +4m^2 >0(4) We look at LHS like as a quadratic polynomial w.r.t “zy” and denote by H(t)(set t=zy,0<zy≤(((z+y)/2))≤1) a)The case 9m^2 +18m+8>0 We consider the discriminant Δ_H ′of H(t) Δ_H ′=(m^2 +2m)^2 +(9m^2 +18m+8)(m^4 −4m^2 ) =m^4 +4m^3 +4m^2 +9m^6 +18m^5 −28m^4 −72m^3 −32m^2 =9m^6 +18m^5 −27m^4 −68m^3 −28m^2 =m^2 (9m^4 +18m^3 −27m^2 −68m−28) <0( due to ∣m∣≤1). Therefore,we infer H(t)>0∀t∈(0;1]which means the inequality (4)is proved ,so f(x)<2 b)The case 9m^2 +18m+8 < 0 ⇔−1<m<−2/3 .We have { ((H(0)=−m^4 +4m^2 =m^2 (4−m^2 )>0 )),((H(1)=−m^4 +15m^2 +22m+8=)),(((1+m)(−m^3 +m^2 +14m+8)>0)) :} By the convert theorem above the sign of the quadratic polynomial we infer H(t)>0 ∀t=zy∈(0;1)⇒(4)is proved which means we ger f(2)<2 other way: Similar to the case i)Rewrite H(t) in the form H(m^2 )as the quadratic poly. w.r.t “m^2 ” with the highest efficient k=(−1)we get H(0)>0,H(1)>0 ⇒kH(0)<0,kH(1)<0⇒H(m)>0 ∀m^2 ∈[0,1]⇔m∈[−1,1] From i)and ii)we obtain 1<f(x)<2(q.e.d)

Q 108815 (19 / 8 / 20) (unanswer) by 1 x . x

Given f (x) = \frac{1}{\sqrt{1 + x}} + \frac{1}{\sqrt{1 + a}} + \frac{\sqrt{ax}}{\sqrt{ax + 8}}

x, a \in R; x, a > 0 . Prove that

1 < f (x) < 2

Solution : Put x = \tan^{2} A, a = \tan^{2} B (A, B \in [0, \frac{π}{2})

f (x) = cosA + cosB + \frac{tanAtanB}{\sqrt{\tan^{2} {Atan}^{2} B + 8}}

= cosA + cosB + \frac{sinAsinB}{\sqrt{{8 \cos}^{2} {Acos}^{2} B + \sin^{2} {Asin}^{2} B}}

Put cosA = z, cosB = y (z, y \in (0, 1]) we have

f = z + y + \frac{\sqrt{(1 - z^{2}) (1 - y^{2})}}{\sqrt{{8 z}^{2} y^{2} + (1 - z^{2}) (1 - y^{2})}}

= z + y + \frac{\sqrt{(1 - z^{2}) (1 - y^{2})}}{\sqrt{{9 z}^{2} y^{2} + 1 - (z^{2} + y^{2})}}

\boldsymbol i) \boldsymbol First \boldsymbol we \boldsymbol prove \boldsymbol f (\boldsymbol x) > 1

\Leftrightarrow z + y + \frac{\sqrt{(1 - z^{2}) (1 - y^{2})}}{\sqrt{{9 z}^{2} y^{2} + 1 - (z^{2} + y^{2})}} > 1 (1)

If z + y ⩾ 1 then the inequality (1) is

true . Consider z + y < 1 . Put

m = 1 - (z + y) \Leftrightarrow z + y = 1 - m (0 < m ⩽ 1)

z^{2} + y^{2} = {(z + y)}^{2} - 2 zy = {(1 - m)}^{2} - 2 zy

(1) \Leftrightarrow \frac{(1 - z^{2}) (1 - y^{2})}{{9 z}^{2} y^{2} + 1 - (z^{2} + y^{2})} > {[1 - (z + y)]}^{2}

\Leftrightarrow 1 + z^{2} y^{2} - (z^{2} + y^{2}) > [{9 z}^{2} y^{2} + 1 - (z^{2} + y^{2})] m^{2}

1 + z^{2} y^{2} - [1 - 2 m + m^{2} - 2 zy] > [{9 z}^{2} y^{2} + 1 - (1 - 2 m + m^{2} - 2 zy)] m^{2}

\Leftrightarrow z^{2} y^{2} + 2 zy + 2 m - m^{2} > ({9 z}^{2} y^{2} + 2 zy + 2 m - m^{2}) m^{2}

\Leftrightarrow m^{4} - m^{2} ({9 z}^{2} y^{2} + 2 zy + 2 m) + (z^{2} y^{2} + 2 zy + 2 m - m^{2}) > 0

We look at LHS as a quadratic polynomial

Prime causes double exponent: use braces to clarify

and we denote by P (m) . By the theorem

above the sign of quadratic poly . P (m) > 0 \Leftrightarrow

Δ_{P} = {({9 z}^{2} y^{2} + 2 zy + 2 m)}^{2} - 4 (z^{2} y^{2} + 2 zy + 2 m - m^{2}) < 0

\Leftrightarrow {81 z}^{4} y^{4} + {4 z}^{2} y^{2} + {4 m}^{2} + {36 z}^{3} y^{3} + {36 mz}^{2} y^{2} + 8 mzy - 4 (z^{2} y^{2} + 2 zy + 2 m - m^{2}) < 0

\Leftrightarrow {81 z}^{4} y^{4} + {36 z}^{3} y^{3} + {36 mz}^{2} y^{2} + 8 (m - 1) zy + {8 m}^{2} - 8 m < 0

\Leftrightarrow {8 m}^{2} + ({36 z}^{2} y^{2} + 8 zy - 8) m + {81 z}^{4} y^{4} + {36 z}^{3} y^{3} - 8 zy < 0 (3)

We look at LHS (3) like as a quadratic

polynomial w . r . t “ m^{″} and denote by Q (m)

We has Q (0) = 81 {(zy)}^{4} + 36 {(zy)}^{3} - 8 zy

⩽ 81 t / 64 + 36 t / 16 - 8 t = 225 t / 64 - 8 t < 0

(due to 0 < t = zy ⩽ 1 / 4)

Q (1) = 1 + 36 {(zy)}^{2} + 8 zy - 8 + 81 {(zy)}^{4} + 36 {(zy)}^{3} - 8 zy

= - 7 + 36 {(zy)}^{2} + 81 {(zy)}^{4} + 36 {(zy)}^{3} ⩽

- 7 + 36 / 16 + 81 / 256 + 36 / 64 < 0 (due to zy ⩽ 1 / 4)

By the convert theorem above the sign

of the quadratic polynomial we infer

Q (m) > 0 \forall m \in (0; 1) which means P (m)

has Δ_{P} < 0 \forall m \in (0; 1) \Rightarrow the inequality (1)

proved

\boldsymbol ii) \boldsymbol Now \boldsymbol we \boldsymbol prove \boldsymbol that \boldsymbol f (\boldsymbol x) < 2

\Leftrightarrow z + y + \frac{\sqrt{(1 - z^{2}) (1 - y^{2})}}{\sqrt{{9 z}^{2} y^{2} + 1 - (z^{2} + y^{2})}} < 2 (4)

\Leftrightarrow \frac{(1 - z^{2}) (1 - y^{2})}{{9 z}^{2} y^{2} + 1 - (z^{2} + y^{2})} < {[2 - (z + y)]}^{2}

\Leftrightarrow 1 + z^{2} y^{2} - (z^{2} + y^{2}) < [{9 z}^{2} y^{2} + 1 - (z^{2} + y^{2})] {(1 + m)}^{2} (note

(1 - m = z + y like as above we have

- 1 ⩽ m = 1 - (z + y) < 1 as 0 < z + y ⩽ 2 (*))

\Leftrightarrow z^{2} y^{2} + 2 zy + 2 m - m^{2} < ({9 z}^{2} y^{2} + 2 zy + 2 m - m^{2}) (1 + 2 m + m^{2})

\Leftrightarrow z^{2} y^{2} + 2 zy + 2 m - m^{2} < ({9 z}^{2} y^{2} + 2 zy + 2 m - m^{2}) + ({9 z}^{2} y^{2} + 2 zy) (2 m + m^{2}) - m^{4} + {4 m}^{2} > 0

\Leftrightarrow - m^{4} + {4 m}^{2} + ({9 z}^{2} y^{2} + 2 zy) (2 m + m^{2}) + {8 z}^{2} y^{2} > 0

\Leftrightarrow ({9 m}^{2} + 18 m + 8) {(zy)}^{2} + 2 (m^{2} + 2 m) zy - m^{4} + {4 m}^{2} > 0 (4)

We look at LHS like as a quadratic

polynomial w . r . t “ {zy}^{″} and denote by

H (t) (set t = zy, 0 < zy ⩽ (\frac{z + y}{2}) ⩽ 1)

\boldsymbol a) \boldsymbol The \boldsymbol case 9 \boldsymbol m^{2} + 18 \boldsymbol m + 8 > 0

We consider the discriminant Δ_{H}^{'} of H (t)

Δ_{H}^{'} = {(m^{2} + 2 m)}^{2} + ({9 m}^{2} + 18 m + 8) (m^{4} - {4 m}^{2})

= m^{4} + {4 m}^{3} + {4 m}^{2} + {9 m}^{6} + {18 m}^{5} - {28 m}^{4} - {72 m}^{3} - {32 m}^{2}

= {9 m}^{6} + {18 m}^{5} - {27 m}^{4} - {68 m}^{3} - {28 m}^{2}

= m^{2} ({9 m}^{4} + {18 m}^{3} - {27 m}^{2} - 68 m - 28)

< 0 (\boldsymbol due \boldsymbol to ∣ \boldsymbol m ∣⩽ 1) . Therefore, we

infer H (t) > 0 \forall t \in (0; 1] which means

the inequality (4) is proved, so f (x) < 2

\boldsymbol b) \boldsymbol The \boldsymbol case 9 \boldsymbol m^{2} + 18 \boldsymbol m + 8 < 0

\Leftrightarrow - 1 < \boldsymbol m < - 2 / 3 . We have

{\begin{cases} H (0) = - m^{4} + {4 m}^{2} = m^{2} (4 - m^{2}) > 0 \\ H (1) = - m^{4} + {15 m}^{2} + 22 m + 8 = \\ (1 + m) (- m^{3} + m^{2} + 14 m + 8) > 0 \end{cases}

By the convert theorem above the sign

of the quadratic polynomial we infer

H (t) > 0 \forall t = zy \in (0; 1) \Rightarrow (4) is proved

which means we ger f (2) < 2

\boldsymbol other \boldsymbol way :

Similar to the case i) Rewrite H (t) in

the form H (m^{2}) as the quadratic poly .

Prime causes double exponent: use braces to clarify

k = (- 1) we get H (0) > 0, H (1) > 0

\Rightarrow kH (0) < 0, kH (1) < 0 \Rightarrow H (m) > 0

\forall m^{2} \in [0, 1] \Leftrightarrow m \in [- 1, 1]

\boldsymbol F \boldsymbol rom \boldsymbol i) \boldsymbol and \boldsymbol ii) \boldsymbol we \boldsymbol obtain 1 < \boldsymbol f (\boldsymbol x) < 2 (\boldsymbol q . \boldsymbol e . \boldsymbol d)

Commented by 1549442205PVT last updated on 21/Aug/20

Thank you, sir . You are welcome

Commented by 1xx last updated on 24/Aug/20

f(x)=(1/( (√(1+x))))+(1/( (√(1+a))))+((√(ax))/( (√(ax+8)))) x>0 , a>0 prove:1<f(x)<2 f^′ (x)=−((1/2))(1/((1+x)(√(1+x))))+(−(1/2))(1/((1+(8/(ax)))(√(1+(8/(ax))))))(8/a)(−(1/x^2 )) f^′ (x)=0⇒(x^2 −(8/a))[x^2 +(8/a)(3−(8/a))x+(8/a)]=0 ∵x>0 ∴x_1 =(√(8/a)) △=[8/a(3−8/a)]^2 −4(8/a)≥0⇒a≤2 f^′ (x)=0⇒ { ((f^′ (x_1 =(√(8/a)))=0 a≥2)),((f^′ (x_1 =(√(8/a)))=f^′ (x_(2,3) )=0 a<2)) :} case A: a≥2 f(x_1 =(√(8/a)))=(2/( (√(1+(√(8/a))))))+(1/( (√(1+a)))) lim_(x→0,∞) f(x)=1+(1/( (√(1+a)))) ∵a≥2 ∴(2/( (√(1+(√(8/a))))))>1 ∴ f_(max) =f(x_1 =(√(8/a))) f(x)>lim_(x→0,∞) f(x)=1+(1/( (√(1+a))))>1 f(x_1 =(√(8/a)))≷2 (2/( (√(1+(√(8/a))))))+(1/( (√(1+a))))≷2 let t=(√(1+(√(8/a)))) then a=(8/((t^2 −1)^2 )) (2/( (√(1+(√(8/a))))))+(1/( (√(1+a))))≷2 ⇔(((t^2 −1)^2 )/((t^2 −1)^2 +8))≷4∙(((t−1)^2 )/t^2 ) ⇔0≷3t^4 −2t^3 −9t^2 +36 ⇔0≷2(t−1)t^3 +(t^2 −6)^2 +3t^2 ∵t>1 ∴(2/( (√(1+(√(8/a))))))+(1/( (√(1+a))))<2 ∴f(x)≤f_(max) =f(x_1 =(√(8/a)))<2 ∵f(x)>lim_(x→0,∞) f(x)=1+(1/( (√(1+a))))>1 ∴ 1<f(x)<2 case B: 0<a<2 f^′ (x_(2,3) )=0 ⇒ x_(2,3) ^2 +(8/a)(3−(8/a))x_(2,3) +(8/a)=0 ⇒x_2 x_3 =(8/a) ∵x_1 =(√(8/a)) ∴x_2 <x_1 <x_3 x_(2,3) ^2 +(8/a)(3−(8/a))x_(2,3) +(8/a)=0 let p=x_(2,3) (ap)^2 +8(3a−8)p+8a=0 ⇒ (ap+8)(ap+a)=(8+a)(ap)+8(8−3a)p ⇒ a(ap+8)(p+1)=(a^2 −16a+64)p ⇒ (1/(1+p))∙((ap)/(ap+8))=(a^2 /((8−a)^2 )) ⇒ (1/( (√(1+x_(2,3) ))))∙((√(ax_(2,3) ))/( (√(ax_(2,3) +8))))=(a/(8−a)) due to 0<a<2 let m=(1/( (√(1+x_(2,3) )))) and n=((√(ax_(2,3) ))/( (√(ax_(2,3) +8)))) then { ((((1/m^2 )−1)((1/n^2 )−1)=(8/a))),((mn=(a/(8−a)))) :} ⇒ (m+n)^2 =1+mn=(8/(8−a)) ⇒ m+n=(√(8/(8−a))) f(x_(2,3) )≷f(x_1 ) ⇔( m+n)≷(2/( (√(1+(√(8/a)))))) ⇔ (√(8/(8−a)))≷(2/( (√(1+(√(8/a)))))) ⇔ ((√8)/( (√(((√8)−(√a))((√8)+(√a))))))≷((2(√(√a)))/( (√(((√8)+(√a)))))) ⇔ (√8)≷(√(4(√a)∙((√8)−(√a)))) ⇔ ((√2))^2 ≷2(√2)(√a)−((√a))^2 ⇔ ((√a)−(√2))^2 ≷0 ∵((√a)−(√2))^2 >0 ∴ f(x_(2,3) )>f(x_1 ) f_(max) =max{f(x_2 ),f(x_3 )}≷2 ⇔( m+n+(1/( (√(1+a)))))≷2 ⇔ (1/2)((√(8/(8−a)))+(1/( (√(1+a)))))≷1 ∵ (1/2)((√(8/(8−a)))+(1/( (√(1+a)))))≤(√((1/2)((8/(8−a)))+(1/(1+a)))) ∴ f_(max) =max{f(x_2 ),f(x_3 )}≷2 ⇔ (√((1/2)((8/(8−a)))+(1/(1+a))))≷1 ⇔ 0≷a(7−2a) ∵ 0<a<2 ∴ 0<a(7−2a) ∴ f_(max) =max{f(x_2 ),f(x_3 )}<2 if f(x_1 )>lim_(x→0,∞) f(x)=1+(1/( (√(1+a))))>1 then 1<f(x)<2 else f_(min) =f(x_1 ) f_(min) =f(x_1 )≷1 ⇔ (2/( (√(1+(√(8/a))))))≷(1−(1/( (√(1+a))))) make Rt△ ((1/( (√(1+a)))))^2 +(((√a)/( (√(1+a)))))^2 =1^2 ∴ ((√a)/( (√(1+a))))>1−(1/( (√(1+a)))) think of (2/( (√(1+(√(8/a))))))≷((√a)/( (√(1+a)))) ⇔ 3a−(√8)(√a)+4≷0 ⇔ 2a+((√a))^2 −2(√2)(√a)+((√2))^2 +2≷0 ⇔ 2a+((√a)−(√2))^2 +2≷0 ∵ a>0 ∴ 2a+((√a)−(√2))^2 +2>0 therefore (2/( (√(1+(√(8/a))))))>((√a)/( (√(1+a))))>1−(1/( (√(1+a)))) f_(min) =f(x_1 )>1 so in case B, 1<f(x)<2 Q.E.D

f (x) = \frac{1}{\sqrt{1 + x}} + \frac{1}{\sqrt{1 + a}} + \frac{\sqrt{a x}}{\sqrt{a x + 8}}

x > 0, a > 0

p r o v e : 1 < f (x) < 2

f^{^{'}} (x) = - (\frac{1}{2}) \frac{1}{(1 + x) \sqrt{1 + x}} + (- \frac{1}{2}) \frac{1}{(1 + \frac{8}{a x}) \sqrt{1 + \frac{8}{a x}}} (8 / a) (- \frac{1}{x^{2}})

f^{^{'}} (x) = 0 \Rightarrow (x^{2} - \frac{8}{a}) [x^{2} + \frac{8}{a} (3 - \frac{8}{a}) x + \frac{8}{a}] = 0

∵ x > 0

∴ x_{1} = \sqrt{8 / a}

△ = {[8 / a (3 - 8 / a)]}^{2} - 4 (8 / a) ⩾ 0 \Rightarrow a ⩽ 2

f^{^{'}} (x) = 0 \Rightarrow {\begin{cases} f^{^{'}} (x_{1} = \sqrt{8 / a}) = 0 a ⩾ 2 \\ f^{^{'}} (x_{1} = \sqrt{8 / a}) = f^{^{'}} (x_{2, 3}) = 0 a < 2 \end{cases}

c a s e A : a ⩾ 2

f (x_{1} = \sqrt{8 / a}) = \frac{2}{\sqrt{1 + \sqrt{\frac{8}{a}}}} + \frac{1}{\sqrt{1 + a}}

\lim_{x \to 0, \infty} f (x) = 1 + \frac{1}{\sqrt{1 + a}}

∵ a ⩾ 2

∴ \frac{2}{\sqrt{1 + \sqrt{\frac{8}{a}}}} > 1

∴ f_{m a x} = f (x_{1} = \sqrt{8 / a})

f (x) > \lim_{x \to 0, \infty} f (x) = 1 + \frac{1}{\sqrt{1 + a}} > 1

f (x_{1} = \sqrt{8 / a}) ≷ 2

\frac{2}{\sqrt{1 + \sqrt{\frac{8}{a}}}} + \frac{1}{\sqrt{1 + a}} ≷ 2

l e t t = \sqrt{1 + \sqrt{\frac{8}{a}}}

t h e n a = \frac{8}{{(t^{2} - 1)}^{2}}

\frac{2}{\sqrt{1 + \sqrt{\frac{8}{a}}}} + \frac{1}{\sqrt{1 + a}} ≷ 2 \Leftrightarrow \frac{{(t^{2} - 1)}^{2}}{{(t^{2} - 1)}^{2} + 8} ≷ 4 \cdot \frac{{(t - 1)}^{2}}{t^{2}}

\Leftrightarrow 0 ≷ 3 t^{4} - 2 t^{3} - 9 t^{2} + 36

\Leftrightarrow 0 ≷ 2 (t - 1) t^{3} + {(t^{2} - 6)}^{2} + 3 t^{2}

∵ t > 1

∴ \frac{2}{\sqrt{1 + \sqrt{\frac{8}{a}}}} + \frac{1}{\sqrt{1 + a}} < 2

∴ f (x) ⩽ f_{m a x} = f (x_{1} = \sqrt{8 / a}) < 2

∵ f (x) > \lim_{x \to 0, \infty} f (x) = 1 + \frac{1}{\sqrt{1 + a}} > 1

∴ 1 < f (x) < 2

c a s e B : 0 < a < 2

f^{^{'}} (x_{2, 3}) = 0 \Rightarrow x_{2, 3}^{2} + \frac{8}{a} (3 - \frac{8}{a}) x_{2, 3} + \frac{8}{a} = 0

\Rightarrow x_{2} x_{3} = \frac{8}{a}

∵ x_{1} = \sqrt{\frac{8}{a}}

∴ x_{2} < x_{1} < x_{3}

x_{2, 3}^{2} + \frac{8}{a} (3 - \frac{8}{a}) x_{2, 3} + \frac{8}{a} = 0

l e t p = x_{2, 3}

{(a p)}^{2} + 8 (3 a - 8) p + 8 a = 0 \Rightarrow

(a p + 8) (a p + a) = (8 + a) (a p) + 8 (8 - 3 a) p \Rightarrow

a (a p + 8) (p + 1) = (a^{2} - 16 a + 64) p \Rightarrow

\frac{1}{1 + p} \cdot \frac{a p}{a p + 8} = \frac{a^{2}}{{(8 - a)}^{2}} \Rightarrow

\frac{1}{\sqrt{1 + x_{2, 3}}} \cdot \frac{\sqrt{{a x}_{2, 3}}}{\sqrt{{a x}_{2, 3} + 8}} = \frac{a}{8 - a} d u e t o 0 < a < 2

l e t m = \frac{1}{\sqrt{1 + x_{2, 3}}}

a n d n = \frac{\sqrt{{a x}_{2, 3}}}{\sqrt{{a x}_{2, 3} + 8}}

t h e n {\begin{cases} (\frac{1}{m^{2}} - 1) (\frac{1}{n^{2}} - 1) = \frac{8}{a} \\ m n = \frac{a}{8 - a} \end{cases} \Rightarrow

{(m + n)}^{2} = 1 + m n = \frac{8}{8 - a} \Rightarrow

m + n = \sqrt{\frac{8}{8 - a}}

f (x_{2, 3}) ≷ f (x_{1}) \Leftrightarrow (m + n) ≷ \frac{2}{\sqrt{1 + \sqrt{\frac{8}{a}}}}

\Leftrightarrow \sqrt{\frac{8}{8 - a}} ≷ \frac{2}{\sqrt{1 + \sqrt{\frac{8}{a}}}}

\Leftrightarrow \frac{\sqrt{8}}{\sqrt{(\sqrt{8} - \sqrt{a}) (\sqrt{8} + \sqrt{a})}} ≷ \frac{2 \sqrt{\sqrt{a}}}{\sqrt{(\sqrt{8} + \sqrt{a})}}

\Leftrightarrow \sqrt{8} ≷ \sqrt{4 \sqrt{a} \cdot (\sqrt{8} - \sqrt{a})}

\Leftrightarrow {(\sqrt{2})}^{2} ≷ 2 \sqrt{2} \sqrt{a} - {(\sqrt{a})}^{2}

\Leftrightarrow {(\sqrt{a} - \sqrt{2})}^{2} ≷ 0

∵ {(\sqrt{a} - \sqrt{2})}^{2} > 0

∴ f (x_{2, 3}) > f (x_{1})

f_{m a x} = m a x {f (x_{2}), f (x_{3})} ≷ 2

\Leftrightarrow (m + n + \frac{1}{\sqrt{1 + a}}) ≷ 2

\Leftrightarrow \frac{1}{2} (\sqrt{\frac{8}{8 - a}} + \frac{1}{\sqrt{1 + a}}) ≷ 1

∵ \frac{1}{2} (\sqrt{\frac{8}{8 - a}} + \frac{1}{\sqrt{1 + a}}) ⩽ \sqrt{\frac{1}{2} (\frac{8}{8 - a}) + \frac{1}{1 + a}}

∴ f_{m a x} = m a x {f (x_{2}), f (x_{3})} ≷ 2

\Leftrightarrow \sqrt{\frac{1}{2} (\frac{8}{8 - a}) + \frac{1}{1 + a}} ≷ 1

\Leftrightarrow 0 ≷ a (7 - 2 a)

∵ 0 < a < 2

∴ 0 < a (7 - 2 a)

∴ f_{m a x} = m a x {f (x_{2}), f (x_{3})} < 2

i f f (x_{1}) > \lim_{x \to 0, \infty} f (x) = 1 + \frac{1}{\sqrt{1 + a}} > 1

t h e n 1 < f (x) < 2

e l s e f_{m i n} = f (x_{1})

f_{m i n} = f (x_{1}) ≷ 1

\Leftrightarrow \frac{2}{\sqrt{1 + \sqrt{\frac{8}{a}}}} ≷ (1 - \frac{1}{\sqrt{1 + a}})

m a k e R t △ {(\frac{1}{\sqrt{1 + a}})}^{2} + {(\frac{\sqrt{a}}{\sqrt{1 + a}})}^{2} = 1^{2}

∴ \frac{\sqrt{a}}{\sqrt{1 + a}} > 1 - \frac{1}{\sqrt{1 + a}}

t h i n k o f \frac{2}{\sqrt{1 + \sqrt{\frac{8}{a}}}} ≷ \frac{\sqrt{a}}{\sqrt{1 + a}}

\Leftrightarrow 3 a - \sqrt{8} \sqrt{a} + 4 ≷ 0

\Leftrightarrow 2 a + {(\sqrt{a})}^{2} - 2 \sqrt{2} \sqrt{a} + {(\sqrt{2})}^{2} + 2 ≷ 0

\Leftrightarrow 2 a + {(\sqrt{a} - \sqrt{2})}^{2} + 2 ≷ 0

∵ a > 0

∴ 2 a + {(\sqrt{a} - \sqrt{2})}^{2} + 2 > 0

t h e r e f o r e \frac{2}{\sqrt{1 + \sqrt{\frac{8}{a}}}} > \frac{\sqrt{a}}{\sqrt{1 + a}} > 1 - \frac{1}{\sqrt{1 + a}}

f_{m i n} = f (x_{1}) > 1

s o i n c a s e B, 1 < f (x) < 2

Q . E . D

Commented by 1xx last updated on 21/Aug/20

Thank to have chance to learn from you. I have finished a proof that is a little bit complex. I think it can be improved by merging your idea.

T h a n k t o h a v e c h a n c e t o l e a r n f r o m y o u .

I h a v e f i n i s h e d a p r o o f t h a t i s a l i t t l e

b i t c o m p l e x . I t h i n k i t c a n b e i m p r o v e d

b y m e r g i n g y o u r i d e a .

Commented by 1xx last updated on 21/Aug/20

we need to confirm : (1+m)(−m^3 +m^2 +14m+8)>0 in case of −1<m<−2/3. when m=−.99, can find (1+m)(−m^3 +m^2 +14m+8)<0

w e n e e d t o c o n f i r m :

(1 + m) (- m^{3} + m^{2} + 14 m + 8) > 0

i n c a s e o f - 1 < m < - 2 / 3 .

w h e n m = - . 99, c a n f i n d

(1 + m) (- m^{3} + m^{2} + 14 m + 8) < 0

Answered by 1xx last updated on 20/Aug/20

in case ii (prove f(x)<2) due to 0<m≤1 ??? pls note: m may be < zero. in case i(prove f(x)>1),we can assume 0<m<1 but it is different in case ii(prove f(x)<2)

i n c a s e i i (p r o v e f (x) < 2)

\boldsymbol due \boldsymbol to 0 < \boldsymbol m ⩽ 1 ? ? ?

p l s n o t e : m m a y b e < z e r o .

i n c a s e i (p r o v e f (x) > 1), w e c a n a s s u m e 0 < m < 1

b u t i t i s d i f f e r e n t i n c a s e i i (p r o v e f (x) < 2)

Commented by 1549442205PVT last updated on 20/Aug/20

Above we limit only consider the z+y<1⇒m=1−(z+y)>0 For ii) Thank you,i missed the case .z+y>1.Please, waiting for i look at again.Corrected

Above we limit only consider the

z + y < 1 \Rightarrow m = 1 - (z + y) > 0

For ii) Thank you, i missed the case

. z + y > 1 . Please,

waiting for i look at again . Corrected

Commented by 1xx last updated on 20/Aug/20

for the same reason, please consider zy<1/4 ??? in case ii(prove f(x)<2)

f o r t h e s a m e r e a s o n, p l e a s e c o n s i d e r

z y < 1 / 4 ? ? ? i n c a s e i i (p r o v e f (x) < 2)

Commented by 1xx last updated on 20/Aug/20

in case of △_H <0, we need to prove 9m^2 +18m+8>0 9m^2 +18m+8=9(m^2 +2m+1)−1 =9(m+1)^2 −1 ⇔∣m+1∣>(1/3) but ∣m∣≤1, only! Q.E.D. is not achieved. Very happy to discuss with you. ps: 0<zy≤(((z^2 +y^2 )/2))≤1 or 0<zy≤(((z+y)/2))^2 ≤1

i n c a s e o f △_{H} < 0, w e n e e d t o p r o v e 9 m^{2} + 18 m + 8 > 0

9 m^{2} + 18 m + 8 = 9 (m^{2} + 2 m + 1) - 1

= 9 {(m + 1)}^{2} - 1

\Leftrightarrow∣ m + 1 ∣> \frac{1}{3}

b u t ∣ m ∣⩽ 1, o n l y!

Q . E . D . i s n o t a c h i e v e d .

V e r y h a p p y t o d i s c u s s w i t h y o u .

p s : 0 < z y ⩽ (\frac{z^{2} + y^{2}}{2}) ⩽ 1 o r 0 < z y ⩽ {(\frac{z + y}{2})}^{2} ⩽ 1

Commented by 1xx last updated on 20/Aug/20

for i)), △_Q >0 need a proof in case of 0<zy<1/4 we should prove (d/dt)(△_Q )=0 have no root in (0,1/4) or we have to consider min(△_Q )∧0

f o r i)), △_{Q} > 0 n e e d a p r o o f i n c a s e o f 0 < z y < 1 / 4

w e s h o u l d p r o v e \frac{d}{d t} (△_{Q}) = 0 h a v e n o r o o t i n (0, 1 / 4)

o r w e h a v e t o c o n s i d e r m i n (△_{Q}) \land 0

Commented by 1549442205PVT last updated on 21/Aug/20

Don′t need to consider Δ because kf(α)<0⇔Δ>0 and x_1 <α<x_2

{Don}^{'} t need to consider Δ because

kf (α) < 0 \Leftrightarrow Δ > 0 and x_{1} < α < x_{2}

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