The discussion revolves around finding the values of $x$ and $y$ given the conditions that $A$, $G$, and $H$ are natural numbers and their sum equals 49. The context involves mathematical reasoning related to inequalities and properties of means.
There appears to be no consensus on the approach to solving for $x$ and $y$, as participants express differing views on the timing of hints and the progression of the discussion.
The discussion lacks specific mathematical steps or assumptions that might clarify the path to finding $x$ and $y$. The conditions under which $A$, $G$, and $H$ are natural numbers may also introduce additional constraints that remain unexplored.
hint :Albert said:we are given :$x,y\in R ,and \,\, x>y>0, A=\dfrac {x+y}{2},G=\sqrt {xy}, H=\dfrac {2xy}{x+y}$
if $A,G,H\in N, \,\, and \,\,\, A+G+H=49$
please find $,x,y$
I hope that you could have given some time before a hint. One day is too short. Here is the solutionAlbert said:hint :
$G^2=A\times H$
$A,G,H $ is a geometric sequence
and $A>G>H$
very good !kaliprasad said:I hope that you could have given some time before a hint. One day is too short. Here is the solution
Let common ratio be r
so $H + Hr + Hr^2= 49$
so $H(1+r+r^2) = 49$
r need not be integer it could be $\dfrac{p}{q}$ but let us try integer
so we get
H is a factor of 49 so H = 1 or 7
H =1 gives r irrational and H= 7 gives r = 2
so $AM = 28$ or $x+ y = 56$ and $xy =196$
solving these 2 as x is greater $x= 28+14\sqrt{3}$ and $y= 28- 14\sqrt{3}$
For r to be fraction h has to be a product of a sqaure that multyiple has to be factor of 49. So h has to be a square or 7 * square
trying $H = 9$ we get $G = 15$ and $A = 25$ and so $x = 45,y = 5$ as the second solution