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Question Number 143410 by mathdanisur last updated on 14/Jun/21

if  x;y;z>0  prove that...  ((x/z))^2 e^(((z/x))^2 ) + ((y/x))^2 e^(((x/y))^2 ) + ((z/y))^2  e^(((y/z))^2 ) ≥ 3e

ifx;y;z>0provethat... (xz)2e(zx)2+(yx)2e(xy)2+(zy)2e(yz)23e

Commented bymr W last updated on 14/Jun/21

using  f(x)=(e^x /x)≥e for x>0

using f(x)=exxeforx>0

Commented bymathdanisur last updated on 14/Jun/21

Sir, exactly how is the solution please

Sir,exactlyhowisthesolutionplease

Commented bymr W last updated on 14/Jun/21

f(x)=(e^x /x)  f′(x)=(e^x /x)−(e^x /x^2 )=0 ⇒x=1  f′′(x)=(e^x /x)−((2e^x )/x^2 )+((2e^x )/x^3 )  f′′(1)=e>0  f(1)=e is minimum of f(x), i.e.  for x>0: (e^x /x)≥e.  ((x/z))^2 e^(((z/x))^2 ) =(e^(((z/x))^2 ) /(((z/x))^2 ))≥e  ((y/x))^2 e^(((x/y))^2 ) =(e^(((x/y))^2 ) /(((x/y))^2 ))≥e  ((z/y))^2 e^(((y/z))^2 ) =(e^(((y/z))^2 ) /(((y/z))^2 ))≥e  ⇒((x/z))^2 e^(((z/x))^2 ) + ((y/x))^2 e^(((x/y))^2 ) + ((z/y))^2  e^(((y/z))^2 ) ≥ e+e+e=3e

f(x)=exx f(x)=exxexx2=0x=1 f(x)=exx2exx2+2exx3 f(1)=e>0 f(1)=eisminimumoff(x),i.e. forx>0:exxe. (xz)2e(zx)2=e(zx)2(zx)2e (yx)2e(xy)2=e(xy)2(xy)2e (zy)2e(yz)2=e(yz)2(yz)2e (xz)2e(zx)2+(yx)2e(xy)2+(zy)2e(yz)2e+e+e=3e

Commented bymathdanisur last updated on 14/Jun/21

cool Sir thank you

coolSirthankyou

Answered by SEIJacob last updated on 14/Jun/21

Answered by mindispower last updated on 14/Jun/21

AM−HM  ⇒  ≥3(((((x^2 .y^2 .z^2 )/(y^2 .z^2 .x^2 )))e^((x^2 /y^2 )+(y^2 /z^2 )+(z^2 /x^2 )) ))^(1/3) =3e^((1/3)((x^2 /y^2 )+(y^2 /z^2 )+(z^2 /x^2 )))   AM−GM  (x^2 /y^2 )+(z^2 /x^2 )+(y^2 /z^2 )≥3(((x^2 .y^2 .z^2 )/(y^2 .z^2 .x^2 )))=3  ⇔≥3e

AMHM 3(x2.y2.z2y2.z2.x2)ex2y2+y2z2+z2x23=3e13(x2y2+y2z2+z2x2) AMGM x2y2+z2x2+y2z23(x2.y2.z2y2.z2.x2)=3 ⇔⩾3e

Commented bymathdanisur last updated on 14/Jun/21

thanks Sir cool

thanksSircool

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