If `27 abc>= (a+b+c)^3 and 3a +4b +5c=12` then `1/a^2+1/b^3+1/c^5=10`, where a, b, c are positive real numbers. Statement-2: For positive real numbers `A.M.>= G.M.`

Options

A. Statement -1 is true, statement-2 is true and statement-2 is correct explanation for statement -1
B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1
C. Statement-1 is true, statement-2 is false
D. Statement-1 is false, statement-2 is true

Answer

To solve the problem, we need to analyze the given statements and equations step by step.

### Given:
1. \( 27abc \geq (a + b + c)^3 \)
2. \( 3a + 4b + 5c = 12 \)
3. We need to check if \( \frac{1}{a^2} + \frac{1}{b^3} + \frac{1}{c^5} = 10 \) holds true for positive real numbers \( a, b, c \).

### Step 1: Analyze Statement 2
The second statement is a known theorem: For positive real numbers, the Arithmetic Mean (AM) is always greater than or equal to the Geometric Mean (GM). Therefore, this statement is true.

**Hint:** Remember that the AM-GM inequality states that for any set of positive numbers, the average (AM) is at least as large as the geometric mean (GM).

### Step 2: Apply AM-GM Inequality
Using the AM-GM inequality on \( a, b, c \):
\[ \frac{a + b + c}{3} \geq (abc)^{1/3} \]
Cubing both sides gives:
\[ \frac{(a + b + c)^3}{27} \geq abc \]
This implies:
\[ (a + b + c)^3 \geq 27abc \]

### Step 3: Equate the Two Inequalities
From the problem, we have:
\[ 27abc \geq (a + b + c)^3 \]
Combining both inequalities:
\[ (a + b + c)^3 \geq 27abc \quad \text{and} \quad 27abc \geq (a + b + c)^3 \]
This means:
\[ (a + b + c)^3 = 27abc \]
The equality in AM-GM holds when \( a = b = c \).

### Step 4: Set \( a = b = c \)
Let \( a = b = c = k \). Then substituting into the second equation:
\[ 3k + 4k + 5k = 12 \implies 12k = 12 \implies k = 1 \]
Thus, \( a = b = c = 1 \).

### Step 5: Substitute Values into the Third Equation
Now, substitute \( a = 1, b = 1, c = 1 \) into the equation:
\[ \frac{1}{a^2} + \frac{1}{b^3} + \frac{1}{c^5} = \frac{1}{1^2} + \frac{1}{1^3} + \frac{1}{1^5} = 1 + 1 + 1 = 3 \]
This does not equal 10.

### Conclusion
Since \( \frac{1}{a^2} + \frac{1}{b^3} + \frac{1}{c^5} = 3 \) does not equal 10, the first statement is false.

### Final Result
- Statement 1 is false.
- Statement 2 is true.

### Summary of Steps:
1. Verified that AM-GM inequality holds true.
2. Used AM-GM to derive that \( (a + b + c)^3 = 27abc \) implies \( a = b = c \).
3. Solved for \( a, b, c \) and substituted back to check the third equation.