Solving x+y+z=2 and 2^{x+y^2}+2^{y+z^2}+2^{z+x^2}=6\sqrt[9]{2} in R

[santa]

solve in R

$$x+y+z=2$$

$$2^{x+y^2}+2^{y+z^2}+2^{z+x^2}=6\sqrt[9]{2}$$

 

[Gib Z]

I don't really want to...

 

[mathman]

solve in R

$$x+y+z=2$$

$$2^{x+y^2}+2^{y+z^2}+2^{z+x^2}=6\sqrt[9]{2}$$

Two equations with three unknowns - usually not possible.

 

[Ben Niehoff]

One solution is x=y=z=2/3.

Since there are only two equations and three unknowns, one can find two of the variables in terms of the third. I'm not sure if the fact that x, y, and z are in R will come into play to eliminate any of the solutions.

 

[sadhu]

One solution is x=y=z=2/3.

Since there are only two equations and three unknowns, one can find two of the variables in terms of the third. I'm not sure if the fact that x, y, and z are in R will come into play to eliminate any of the solutions.

you are absolutely right in your answer but that requires a good assumption that x=y=z

however later on working out the question myself i found that you assump. was right and can be prooved
here
given
x+y+z=2
2$$^{x+x^2}$$+2$$^{y+y^2}$$ +2$$^{z+z^2}$$=6*2$$^{10/9}$$a,b,c are the given terms in equation
using AM GM inequality in a,b,c
2$$^{10/9}$$>=2$$^{(x+y+z+x^2+y^2+z^2)/3}$$
x$$^{2}$$+y$$^{2}$$+z$$^{2}$$<=4/3
now

x+y+z=2
x$$^{2}$$+y$$^{2}$$+z$$^{2}$$=$$\lambda$$

make an equation containing the solutions
let it is in variable a
a$$^{3}$$-2*a$$^{2}$$+((4-$$\lambda$$)/2)*a-k=0
differentiate the equation w.r.t a
3*a$$^{2}$$-4*a+((4-$$\lambda$$)/2)
for three distinct real sol. of equation
$$\Delta$$>0
$$\Delta$$=16-24+6$$\lambda$$>0
$$\lambda$$>4/3

similiarly
$$\Delta$$=0...(2,1 no of sol.)
$$\Delta$$<0...(only one possible)

therefore we get
x$$^{2}$$+y$$^{2}$$+z$$^{2}$$>4/3 ...for 3,2,1 no of distinct sol.
x$$^{2}$$+y$$^{2}$$+z$$^{2}$$=4/3...for(2,1 no of sol.)
x$$^{2}$$+y$$^{2}$$+z$$^{2}$$<4/3...(for 1 sol i.e x=y=z)
...by second process
x$$^{2}$$+y$$^{2}$$+z$$^{2}$$<=4/3.....by first processnow you can see that only condition that satisfy both equations are

x$$^{2}$$+y$$^{2}$$+z$$^{2}$$=4/3
if
x$$^{2}$$+y$$^{2}$$+z$$^{2}$$<4/3
then that would be against AM-GM
in AM GM inequality
equality is only possible when all terms are equal
a=b=c
or
x=y=z=k
3k=2
k=2/3

hope this is correct

 

[Defennder]

make an equation containing the solutions
let it is in variable a
a$$^{3}$$-2*a$$^{2}$$+((4-$$\lambda$$)/2)*a-k=0
differentiate the equation w.r.t a
3*a$$^{2}$$-4*a+((4-$$\lambda$$)/2)
for three real sol. of equation
$$\Delta$$>=0
$$\Delta$$=16-24+6$$\lambda$$>=0
$$\lambda$$>=4/3

You lost me here. I don't follow what you mean by the above, where does $$a^3 - 2a^2 + \frac{a(4-\lambda)}{2} - k = 0$$ come from?

More importantly what does this mean?

a,b,c are the given terms in equation
using AM GM inequality in a,b,c

You don't appear to have used b,c at all. And I thought you used the AM-GM inequality for variables x,y,z only. Where does b,c come into play?

 

[sadhu]

You lost me here. I don't follow what you mean by the above, where does $$a^3 - 2a^2 + \frac{a(4-\lambda)}{2} - k = 0$$ come from?

More importantly what does this mean?

You don't appear to have used b,c at all. And I thought you used the AM-GM inequality for variables x,y,z only. Where does b,c come into play?

assume an equation
(a-x)(a-y)(a-z)=0
now all values of x,y,z are present in the solution of the equation
put the coefficients for expansion accordindly and you get the above equation

 

[sadhu]

while using AM GM inequality
a,b,c
represent the terms of the equation ,not x,y,z
however the answer after manipulation comes out in terms of x,y,z

(2$$^{x+x^2}$$+2$$^{z+z^2}$$+2$$^{y+y^2}$$)/3>=2$$^{(x+x^2+y+y^2+z+z^2)/3}$$
substitute the x+y+z=2 in GM & equation inAM term
and solve indices equation
you will get the equation you want

 

[Defennder]

assume an equation
(a-x)(a-y)(a-z)=0
now all values of x,y,z are present in the solution of the equation
put the coefficients for expansion accordindly and you get the above equation

Actually I don't get that equation. I get $$a^3 - 2a^2 + a(xy+z(x+y)) - xyz = 0$$

Secondly, why do you differentiate the equation? How is the solution for 'a' to the differentiated equation the same as that to the original equation for 'a'?

 

[sadhu]

you got the correct equation
x+y+z=2

{(x+y+z)^2-(x^2+y^2+z^2)}/2=xy+yz+xzsubstituting x+y+z=2 and
x$$^{2}$$+y$$^{2}$$+z$$^{2}$$=$$\lambda$$
we get

(4-lambda)/2

take xyz as any number let it be k
since it doesn't matter the solutionnow you see it is given that x,y,z are real numbers
suppose that they have three , two distinct value(just suppose)so that means that a - curve must cut the x-axis on three or two points...

as you know that if any graph cut x-axis on n point then it must have (n-1) (min. no )solutions of maxima and minima
for three-2
for 2-1 ....these are minimum no. of maxima or minima sol.
for-0
actualy up till now we cannot say that whether it is 2 or1or3

but as you can see when you differentiate the equation and solve for maxima and minima
if $$\Delta$$>0 then two solutions are there and you may get three distinct value of a,or two values or one value
but
$$\Delta$$=0
then you may get two or one solution of a
however you know that two solution cannot be their in this case as this option is eliminated by AM -GM inequality which on equality demand x=y=z
third possibility is
$$\Delta$$<0
that means no maxima or minima, that again mean that their can be only one solution to a or x=y=z
but the reason i did not mention it was because
it will form

x$$^{2}$$+y$$^{2}$$+z$$^{2}$$<4/3
but it also would mean that if this is possible then x=y=z

but AM-GM contradict it because if

x$$^{2}$$+y$$^{2}$$+z$$^{2}$$<4/3
then x!=y!=z

therefore only $$\Delta$$=0 is satisfied...hence i did not mention it one in my earlier post to avoid confusion

sorry for that...-:)
hope you get it

 

[Defennder]

as you know that if any graph cut x-axis on n point then it must have (n-1) (min. no )solutions of maxima and minima
for three-2
for 2-1 ....these are minimum no. of maxima or minima sol.
for-0
actualy up till now we cannot say that whether it is 2 or1or3

Where can I find this theorem? I could find only the apparently trivial result that if f(a) and f(b) are of opposite sign then a root exists in the interval [a,b]. And how is it the case that if $$\Delta = 0$$ there can be only 1 solution for x,y,z? Shouldn't it be 2 by your quote above? I can understand everything else except this point.

 

[sadhu]

well i don't care about theorems or methods or what ever?

I don't know whether it is really some theorem or not , i only know that this is a result apparently
coming out of graphs ...only need is that the function should be differentiable at all points in domain.
however on doing some search in some books, i found that this result is called rolles theorem in many textbooks .

 

[sadhu]

when
$$\Delta$$=0
there can be two or one solution , as we get from the graphs

but it means that$$\lambda$$=0
i.e
x$$^{2}$$+y$$^{2}$$+z$$^{2}$$=0

when we see the AM-GM inequality
if
x$$^{2}$$+y$$^{2}$$+z$$^{2}$$=0
then x=y=z,and only this condition
hence (x,y,z) they can't have 2 different values distributed amongst them
hence there can't be two solutions to the equation in a, as this is going against the deduction of AM-GM


hope you get it

also i think that someone will come up with a better solution than this

 

[Defennder]

Thanks for your explanation. I think your method is rather brilliant, but then it must follow that there being only 1 solution does not strictly imply $$\Delta < 0$$, right?

 

[sadhu]

you are right , the observation talks about minimum no of solution needed
when 1,
we get by formula 1-1=0
hence this is minimum we require for 1 solution ,i.e it is always possibleexample
consider a function having
2 max,min
but all are far below or above the x-axis

you can see that only one solution is coming out ,however
now you can see that$$\Delta$$>0however in this case we get one solution by $$\Delta$$=0
as other solutions are negated by disparity between two inequalities obtained by two different processes.

 

[Defennder]

Hmm ok thanks a lot. By the way I don't think that theorem is known as Rolle's theorem, but I couldn't seem to find a proof of it. I searched the calculus texts in the library earlier today and the closest I got was that if f(x) is continuous in an interval and in f(a) and f(b) have opposite signs then f(x) crosses the x-axis in the interval an odd number of times.

 

[sadhu]

that is again one very obvious conclusion , you can make from just seeing the graphs, and just thinking a bit.

well i think that one can still use the results which are very obvious even without going into a rigorous mathematical proof , which might be far more difficult than the result itself...

this result is very obvious once you know rolles theorem

as if, there are 3 solution then there are two intervals and each interval correspond
to the solution of derivative=0.....by rolles theorem
thus max,min are 2 corresponding to two intervals

hence the deduction is proved

 

[Defennder]

Hm ok thanks