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Question Number 142986 by mnjuly1970 last updated on 08/Jun/21

Commented by MJS_new last updated on 08/Jun/21

if y=x the expression is constantly  (4)^(1/3)

ify=xtheexpressionisconstantly43

Commented by mnjuly1970 last updated on 08/Jun/21

perfec...

perfec...

Answered by TheHoneyCat last updated on 17/Jun/21

ler c: { (R,→,R),(x,→,x^3 ) :} ∈C^∞ (R) which, by checking the derrivative, is obviously a bijection  So:  Min_(x,y>1) {((^3 (√(x−1))+^3 (√(y−1)))/(^3 (√(x+y−2))))}  =Min_(x,y>0) {((^3 (√x)+^3 (√y))/(^3 (√(x+y))))} _((x⇋x−1, y⇋y−1))   =c^(−1) (Min_(x,y>0) {(((^3 (√x)+^3 (√y))^3 )/(x+y))})   =c^(−1) (Min_(x,y>0) {(((x+y)^3 )/(x^3 +y^3 ))}) _((x⇋^3 (√x), y⇋^3 (√y)))   =c^(−1) (Min_(x,y>0) {((x^3 +y^3 +6xy^2 +6yx^2 )/(x^3 +y^3 ))})  =c^(−1) (Min_(x,y>0) {1+6((xy^2 +yx^2 )/(x^3 +y^3 ))})  =c^(−1) (1+6×Min_(x,y>0) {((xy^2 +yx^2 )/(x^3 +y^3 ))})    let (Un)=(1,(1/n))_(n∈N)   (Un)∈{x,y ∣ x>0 and y>0} = ′′x,y>0′′  and (((1/n^2 )+(1/n))/(1+(1/n^3 )))→_(n→+∞) 0  So, Min_(x,y>0) {((xy^2 +yx^2 )/(x^3 +y^3 ))}≤0  But ∀(x,y)∈(R_+ ^∗ )^2  ((xy^2 +yx^2 )/(x^3 +y^3 ))}≥0 (∗)  So, Min_(x,y>0) {((xy^2 +yx^2 )/(x^3 +y^3 ))}=0    And thus :  Min_(x,y>1) {((^3 (√(x−1))+^3 (√(y−1)))/(^3 (√(x+y−2))))}=c^(−1) (1)=1  Your minimum is worth 1    By the way, it is possible to proove that this value cannot be obtained with (x,y)∈(R_+ ^∗ )^2   (∗) is in fact a strict innequality  so technically it is not called a Max but a Sup  and (∗) would not contradict anything.

lerc:{RRxx3C(R)which,bycheckingthederrivative,isobviouslyabijectionSo:Minx,y>1{3x1+3y13x+y2}=Minx,y>0{3x+3y3x+y}(xx1,yy1)=c1(Minx,y>0{(3x+3y)3x+y})=c1(Minx,y>0{(x+y)3x3+y3})(x3x,y3y)=c1(Minx,y>0{x3+y3+6xy2+6yx2x3+y3})=c1(Minx,y>0{1+6xy2+yx2x3+y3})=c1(1+6×Minx,y>0{xy2+yx2x3+y3})let(Un)=(1,1n)nN(Un){x,yx>0andy>0}=x,y>0and1n2+1n1+1n3n+0So,Minx,y>0{xy2+yx2x3+y3}0But(x,y)(R+)2xy2+yx2x3+y3}0()So,Minx,y>0{xy2+yx2x3+y3}=0Andthus:Minx,y>1{3x1+3y13x+y2}=c1(1)=1Yourminimumisworth1Bytheway,itispossibletoproovethatthisvaluecannotbeobtainedwith(x,y)(R+)2()isinfactastrictinnequalitysotechnicallyitisnotcalledaMaxbutaSupand()wouldnotcontradictanything.

Answered by ajfour last updated on 17/Jun/21

x−1=p^3  , y−1=q^3   z=((p+q)/((p^3 +q^3 )^(1/3) ))=((m+1)/((m^3 +1)^(1/3) ))  z=(((m+1)^(2/3) )/((m^2 −m+1)^(1/3) ))  z^3 =(((m+1)^2 )/(m^2 −m+1))=(((m+1)^2 )/((m+1)^2 −3(m+1)+3))    let  ((m+1)/( (√3)))=t  ⇒ z^3 =((3t^2 )/(3t^2 −3(√3)t+3))  z^3 =(t/(t+(1/t)−(√3)))  say t+(1/t)=s ⇒ t=(s/2)±(√((s^2 /4)−1))  z^3 =((s−(√(s^2 −4)))/(2(s−(√3))))  ((d(2z^3 ))/ds)=(((s−(√3))(1−(s/( (√(s^2 −4)))))−(s−(√(s^2 −4))))/((s−(√3))^2 ))  ((d(2z^3 ))/ds)=(((((√3)s−s^2 )/( (√(s^2 −4))))−(√3)+(√(s^2 −4)))/((s−3(√3))^2 ))       =(((√3)s−4−(√3)(√(s^2 −4)))/( (√(s^2 −4))(s−3(√3))^2 ))  let   ((√3)s−4)^2 =3(s^2 −4)  ⇒ s=(7/( 2(√3)))  z^3 =(((((7−1)/( 2(√3)))))/(((2/( 2(√3))))))=3  z=(3)^(1/3)

x1=p3,y1=q3z=p+q(p3+q3)1/3=m+1(m3+1)1/3z=(m+1)2/3(m2m+1)1/3z3=(m+1)2m2m+1=(m+1)2(m+1)23(m+1)+3letm+13=tz3=3t23t233t+3z3=tt+1t3sayt+1t=st=s2±s241z3=ss242(s3)d(2z3)ds=(s3)(1ss24)(ss24)(s3)2d(2z3)ds=3ss2s243+s24(s33)2=3s43s24s24(s33)2let(3s4)2=3(s24)s=723z3=(7123)(223)=3z=33

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