Statement-1: If x, y are positive real number satisfying `x+y=1`, then `x^(1//3)+y^(1//3)gt2^(2//3)`
Statement-2: `(x^(n)+y^(n))/(2)lt((x+y)/(2))^(n)`, if `0ltnlt1` and `x,ygt0.`

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to analyze both statements provided and determine their validity.

### Step 1: Analyze Statement 2
Statement 2 states that:
\[ \frac{x^n + y^n}{2} < \left(\frac{x+y}{2}\right)^n \]
for \(0 < n < 1\) and \(x, y > 0\).

**Proof:**
This statement is a known result from the Power Mean Inequality. The Power Mean Inequality states that for any two positive numbers \(x\) and \(y\) and for \(p < q\):
\[ \left(\frac{x^p + y^p}{2}\right)^{1/p} < \left(\frac{x^q + y^q}{2}\right)^{1/q} \]
In our case, we can set \(p = n\) and \(q = 1\). Since \(0 < n < 1\), we have:
\[ \frac{x^n + y^n}{2} < \left(\frac{x+y}{2}\right)^n \]
Thus, Statement 2 is **true**.

### Step 2: Analyze Statement 1
Statement 1 states that:
\[ x^{1/3} + y^{1/3} > 2^{2/3} \]
given that \(x + y = 1\) and \(x, y > 0\).

**Proof:**
Using the result from Statement 2, we can apply it with \(n = \frac{1}{3}\):
\[ \frac{x^{1/3} + y^{1/3}}{2} < \left(\frac{x+y}{2}\right)^{1/3} \]
Since \(x + y = 1\), we have:
\[ \frac{x^{1/3} + y^{1/3}}{2} < \left(\frac{1}{2}\right)^{1/3} \]
Multiplying both sides by 2:
\[ x^{1/3} + y^{1/3} < 2 \cdot \left(\frac{1}{2}\right)^{1/3} \]
Calculating the right side:
\[ 2 \cdot \left(\frac{1}{2}\right)^{1/3} = 2^{1 - \frac{1}{3}} = 2^{\frac{2}{3}} \]
Thus, we have:
\[ x^{1/3} + y^{1/3} < 2^{2/3} \]
This contradicts Statement 1, which claims that \(x^{1/3} + y^{1/3} > 2^{2/3}\). Therefore, Statement 1 is **false**.

### Conclusion
- Statement 1 is **false**.
- Statement 2 is **true**.

Thus, the final answer is that Statement 1 is incorrect and Statement 2 is correct.

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