The discussion focuses on solving for variables $x$ and $y$ given the equations involving arithmetic mean $A$, geometric mean $G$, and harmonic mean $H$. Specifically, the conditions state that $A = \frac{x+y}{2}$, $G = \sqrt{xy}$, $H = \frac{2xy}{x+y}$, and that all three means must be natural numbers summing to 49. The solution involves manipulating these equations to find valid pairs of $(x, y)$ that satisfy the conditions of being real numbers with $x > y > 0$.
PREREQUISITESMathematics students, educators, and anyone interested in problem-solving involving means and inequalities will benefit from this discussion.
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