Menu Close

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)
Q108815(19/8/20)(unanswer)by1x.xGivenf(x)=11+x+11+a+axax+8x,aR;x,a>0.Provethat1<f(x)<2Solution:Putx=tan2A,a=tan2B(A,B[0,π2)f(x)=cosA+cosB+tanAtanBtan2Atan2B+8=cosA+cosB+sinAsinB8cos2Acos2B+sin2Asin2BPutcosA=z,cosB=y(z,y(0,1])wehavef=z+y+(1z2)(1y2)8z2y2+(1z2)(1y2)=z+y+(1z2)(1y2)9z2y2+1(z2+y2)i)Firstweprovef(x)>1z+y+(1z2)(1y2)9z2y2+1(z2+y2)>1(1)Ifz+y1thentheinequality(1)istrue.Considerz+y<1.Putm=1(z+y)z+y=1m(0<m1)z2+y2=(z+y)22zy=(1m)22zy(1)(1z2)(1y2)9z2y2+1(z2+y2)>[1(z+y)]21+z2y2(z2+y2)>[9z2y2+1(z2+y2)]m21+z2y2[12m+m22zy]>[9z2y2+1(12m+m22zy)]m2z2y2+2zy+2mm2>(9z2y2+2zy+2mm2)m2m4m2(9z2y2+2zy+2m)+(z2y2+2zy+2mm2)>0WelookatLHSasaquadraticpolynomialPrime causes double exponent: use braces to clarifyandwedenotebyP(m).Bythetheoremabovethesignofquadraticpoly.P(m)>0ΔP=(9z2y2+2zy+2m)24(z2y2+2zy+2mm2)<081z4y4+4z2y2+4m2+36z3y3+36mz2y2+8mzy4(z2y2+2zy+2mm2)<081z4y4+36z3y3+36mz2y2+8(m1)zy+8m28m<08m2+(36z2y2+8zy8)m+81z4y4+36z3y38zy<0(3)WelookatLHS(3)likeasaquadraticpolynomialw.r.tmanddenotebyQ(m)WehasQ(0)=81(zy)4+36(zy)38zy81t/64+36t/168t=225t/648t<0(dueto0<t=zy1/4)Q(1)=1+36(zy)2+8zy8+81(zy)4+36(zy)38zy=7+36(zy)2+81(zy)4+36(zy)37+36/16+81/256+36/64<0(duetozy1/4)BytheconverttheoremabovethesignofthequadraticpolynomialweinferQ(m)>0m(0;1)whichmeansP(m)hasΔP<0m(0;1)theinequality(1)provedii)Nowweprovethatf(x)<2z+y+(1z2)(1y2)9z2y2+1(z2+y2)<2(4)(1z2)(1y2)9z2y2+1(z2+y2)<[2(z+y)]21+z2y2(z2+y2)<[9z2y2+1(z2+y2)](1+m)2(note(1m=z+ylikeasabovewehave1m=1(z+y)<1as0<z+y2())z2y2+2zy+2mm2<(9z2y2+2zy+2mm2)(1+2m+m2)z2y2+2zy+2mm2<(9z2y2+2zy+2mm2)+(9z2y2+2zy)(2m+m2)m4+4m2>0m4+4m2+(9z2y2+2zy)(2m+m2)+8z2y2>0(9m2+18m+8)(zy)2+2(m2+2m)zym4+4m2>0(4)WelookatLHSlikeasaquadraticpolynomialw.r.tzyanddenotebyH(t)(sett=zy,0<zy(z+y2)1)a)Thecase9m2+18m+8>0WeconsiderthediscriminantΔHofH(t)ΔH=(m2+2m)2+(9m2+18m+8)(m44m2)=m4+4m3+4m2+9m6+18m528m472m332m2=9m6+18m527m468m328m2=m2(9m4+18m327m268m28)<0(duetom∣⩽1).Therefore,weinferH(t)>0t(0;1]whichmeanstheinequality(4)isproved,sof(x)<2b)Thecase9m2+18m+8<01<m<2/3.Wehave{H(0)=m4+4m2=m2(4m2)>0H(1)=m4+15m2+22m+8=(1+m)(m3+m2+14m+8)>0BytheconverttheoremabovethesignofthequadraticpolynomialweinferH(t)>0t=zy(0;1)(4)isprovedwhichmeanswegerf(2)<2otherway:Similartothecasei)RewriteH(t)intheformH(m2)asthequadraticpoly.Prime causes double exponent: use braces to clarifyk=(1)wegetH(0)>0,H(1)>0kH(0)<0,kH(1)<0H(m)>0m2[0,1]m[1,1]Fromi)andii)weobtain1<f(x)<2(q.e.d)
Commented by 1549442205PVT last updated on 21/Aug/20
Thank you,sir.You are welcome
Thankyou,sir.Youarewelcome
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)=11+x+11+a+axax+8x>0,a>0prove:1<f(x)<2f(x)=(12)1(1+x)1+x+(12)1(1+8ax)1+8ax(8/a)(1x2)f(x)=0(x28a)[x2+8a(38a)x+8a]=0x>0x1=8/a=[8/a(38/a)]24(8/a)0a2f(x)=0{f(x1=8/a)=0a2f(x1=8/a)=f(x2,3)=0a<2caseA:a2f(x1=8/a)=21+8a+11+alimx0,f(x)=1+11+aa221+8a>1fmax=f(x1=8/a)f(x)>limx0,f(x)=1+11+a>1f(x1=8/a)221+8a+11+a2lett=1+8athena=8(t21)221+8a+11+a2(t21)2(t21)2+84(t1)2t203t42t39t2+3602(t1)t3+(t26)2+3t2t>121+8a+11+a<2f(x)fmax=f(x1=8/a)<2f(x)>limx0,f(x)=1+11+a>11<f(x)<2caseB:0<a<2f(x2,3)=0x2,32+8a(38a)x2,3+8a=0x2x3=8ax1=8ax2<x1<x3x2,32+8a(38a)x2,3+8a=0letp=x2,3(ap)2+8(3a8)p+8a=0(ap+8)(ap+a)=(8+a)(ap)+8(83a)pa(ap+8)(p+1)=(a216a+64)p11+papap+8=a2(8a)211+x2,3ax2,3ax2,3+8=a8adueto0<a<2letm=11+x2,3andn=ax2,3ax2,3+8then{(1m21)(1n21)=8amn=a8a(m+n)2=1+mn=88am+n=88af(x2,3)f(x1)(m+n)21+8a88a21+8a8(8a)(8+a)2a(8+a)84a(8a)(2)222a(a)2(a2)20(a2)2>0f(x2,3)>f(x1)fmax=max{f(x2),f(x3)}2(m+n+11+a)212(88a+11+a)112(88a+11+a)12(88a)+11+afmax=max{f(x2),f(x3)}212(88a)+11+a10a(72a)0<a<20<a(72a)fmax=max{f(x2),f(x3)}<2iff(x1)>limx0,f(x)=1+11+a>1then1<f(x)<2elsefmin=f(x1)fmin=f(x1)121+8a(111+a)makeRt(11+a)2+(a1+a)2=12a1+a>111+athinkof21+8aa1+a3a8a+402a+(a)222a+(2)2+202a+(a2)2+20a>02a+(a2)2+2>0therefore21+8a>a1+a>111+afmin=f(x1)>1soincaseB,1<f(x)<2Q.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.
Thanktohavechancetolearnfromyou.Ihavefinishedaproofthatisalittlebitcomplex.Ithinkitcanbeimprovedbymergingyouridea.
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
weneedtoconfirm:(1+m)(m3+m2+14m+8)>0incaseof1<m<2/3.whenm=.99,canfind(1+m)(m3+m2+14m+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)
incaseii(provef(x)<2)dueto0<m1???plsnote:mmaybe<zero.incasei(provef(x)>1),wecanassume0<m<1butitisdifferentincaseii(provef(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
Abovewelimitonlyconsiderthez+y<1m=1(z+y)>0Forii)Thankyou,imissedthecase.z+y>1.Please,waitingforilookatagain.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)
forthesamereason,pleaseconsiderzy<1/4???incaseii(provef(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
incaseofH<0,weneedtoprove9m2+18m+8>09m2+18m+8=9(m2+2m+1)1=9(m+1)21⇔∣m+1∣>13butm∣⩽1,only!Q.E.D.isnotachieved.Veryhappytodiscusswithyou.ps:0<zy(z2+y22)1or0<zy(z+y2)21
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
fori)),Q>0needaproofincaseof0<zy<1/4weshouldproveddt(Q)=0havenorootin(0,1/4)orwehavetoconsidermin(Q)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
DontneedtoconsiderΔbecausekf(α)<0Δ>0andx1<α<x2

Leave a Reply

Your email address will not be published. Required fields are marked *