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Sequences & Series: Positive Numbers Minimum Value Less

JEE Maths question with a full step-by-step solution.

Question

If positive numbers

x

y

z

are in A.P., then the minimum value of

\frac{x+y}{2y-x} + \frac{y+z}{2y-z}

is less than

2

6

correct

8

correct

1

Solution

Step 1: Use the A.P. condition to simplify the two denominators. Since

x, y, z

are in A.P.,

2y = x + z .

Therefore

2y - x = z \qquad \text{and} \qquad 2y - z = x .

Step 2: Rewrite the expression - it is much friendlier now.

E = \frac{x+y}{z} + \frac{y+z}{x} .

Step 3: Replace

y

\dfrac{x+z}{2}

E = \frac{x + \dfrac{x+z}{2}}{z} + \frac{\dfrac{x+z}{2} + z}{x} = \frac{3x+z}{2z} + \frac{x+3z}{2x} .

Step 4: Split each fraction into its pieces.

E = \frac{3x}{2z} + \frac{1}{2} + \frac{1}{2} + \frac{3z}{2x} = 1 + \frac32\left(\frac{x}{z} + \frac{z}{x}\right).

Step 5: Apply AM

\ge

GM to

\dfrac{x}{z}

and

\dfrac{z}{x}

, both positive:

\frac{x}{z} + \frac{z}{x} \ge 2 .

Step 6: Hence

E \ge 1 + \frac32 \times 2 = 4 ,

with equality when

x = z

(and then

y = x

too, so all three are equal). Step 7: The minimum value is

4

. Compare with the options:

4

is not less than

1

and

2

, but it is less than

6

and less than

8

. Answer: (2) and (3).

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