Question

Which statement is correct?
(A) \[\dfrac{2x+3y}{5}\ge 3\ge \dfrac{5}{\dfrac{3}{x}+\dfrac{2}{y}}\]
(B) \[\dfrac{2x+3y}{5}\ge 3\ge \dfrac{5xy}{3x+2y}\]
(C) \[\dfrac{2x+3y}{5}\ge 3\ge \dfrac{5xy}{3x+4y}\]
(D) \[\dfrac{2x+3y}{5}\ge 3\ge \dfrac{5xy}{2x+3y}\]

Answer 1

Hint: Assume that we have two real positive numbers x and y such that \[x=y=3\] . We know the property that for any real positive numbers, \[A.M\ge G.M\ge H.M\] . Now, calculate the arithmetic mean of the real positive numbers \[x,x,y,y\] , and \[y\] . The arithmetic mean of the real positive numbers \[x,x,y,y\] and \[y\] should be equal to \[\dfrac{2x+3y}{5}\] . Then, calculate the geometric mean of the real positive numbers \[x,x,y,y\] , and \[y\] . The geometric mean of the real positive numbers \[x,x,y,y\] and \[y\] should be equal to \[{{\left( {{x}^{2}}{{y}^{3}} \right)}^{\dfrac{1}{5}}}\] . Now, put \[x=y=3\] and simplify \[{{\left( {{x}^{2}}{{y}^{3}} \right)}^{\dfrac{1}{5}}}\] . Now, calculate the harmonic mean of the real positive numbers \[x,x,y,y\] , and \[y\] . The harmonic mean of the real positive numbers \[x,x,y,y\] and \[y\] should be equal to \[\dfrac{5}{\dfrac{2}{x}+\dfrac{3}{y}}\] . We know the property that for any real positive numbers, \[A.M\ge G.M\ge H.M\] . Apply this property for the numbers \[x,x,y,y\] and \[y\] . Now, solve it further and get the required answer.

Complete step by step answer:

First of all, let us assume that we have two real positive numbers x and y such that \[x=y=3\] .
The value of x = 3 ………………………………………….(1)
The value of y = 3 …………………………………………(2)
We know the property that for any real positive numbers, \[A.M\ge G.M\ge H.M\] . Here, A.M, G.M, and H.M are arithmetic mean, geometric mean, and harmonic mean of the numbers …………………………………(3)
In the options given above, we can see that the coefficient of x is 2. It means that the number x is added twice. Similarly, we can also see that the coefficient of y is 3. It means that the number y is added thrice.
Now, take the real positive numbers \[x,x,y,y,y\] .
The arithmetic mean of the above numbers = \[\dfrac{x+x+y+y+y}{5}=\dfrac{2x+3y}{5}\] ………………………………………(4)
The geometric mean of the above numbers = \[{{\left( x.x.y.y.y \right)}^{\dfrac{1}{5}}}={{\left( {{x}^{2}}{{y}^{3}} \right)}^{\dfrac{1}{5}}}\] ………………………………………(5)
From equation (1) and equation (2), we have the value of x and y.
Now, on putting \[x=3\] and \[y=3\] in equation (5), we get

The geometric mean of the above numbers = \[{{\left( {{3}^{2}}{{3}^{3}} \right)}^{\dfrac{1}{5}}}={{\left( {{3}^{5}} \right)}^{\dfrac{1}{5}}}\] …………………………………(6)
The harmonic mean of the above numbers = \[\dfrac{5}{\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{y}}=\dfrac{5}{\dfrac{2}{x}+\dfrac{3}{y}}\] ………………………………………(7)
Now, on applying the property shown in equation (3) for the numbers \[x,x,y,y\], and \[y\] , we get
\[A.M\ge G.M\ge H.M\]
\[\begin{align}
  & \dfrac{2x+3y}{5}\ge {{\left( {{3}^{5}} \right)}^{\dfrac{1}{5}}}\ge \dfrac{5}{\dfrac{2}{x}+\dfrac{3}{y}} \\
 & \dfrac{2x+3y}{5}\ge 3\ge \dfrac{5}{\dfrac{2}{x}+\dfrac{3}{y}} \\
 & \dfrac{2x+3y}{5}\ge 3\ge \dfrac{5}{\dfrac{2y+3x}{xy}} \\
 & \dfrac{2x+3y}{5}\ge 3\ge \dfrac{5xy}{3x+2y} \\
\end{align}\]
Therefore, \[\dfrac{2x+3y}{5}\ge 3\ge \dfrac{5xy}{3x+2y}\] .
Hence, the correct option is (B).

Note:
In this question, one might try to verify each option by putting random real values of x and y in the options given. But this approach is not going to work here because it is very complex to put every real value of x and y and verify the options given. Therefore, the best way to approach this question is by using the property that for any real positive numbers, \[A.M\ge G.M\ge H.M\].