Let `a,b,c,d in R` such that `a^2+b^2+c^2+d^2=25`

Options

A. `ab+bc+c d+dale(25)/(2)`
B. `ab+bc+c d+dale25`
C. `ab+bc+c d+dale5`
D. `ab+bc+c d+dale(5)/(2)`

Explanation

To solve the problem, we need to analyze the given condition \( a^2 + b^2 + c^2 + d^2 = 25 \) and apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality to derive relationships between the variables.

### Step-by-Step Solution:

1. **Apply AM-GM Inequality on \( a^2 \) and \( b^2 \)**:
\[ \frac{a^2 + b^2}{2} \geq \sqrt{a^2 b^2} = ab \]
This implies:
\[ a^2 + b^2 \geq 2ab \tag{1} \]

2. **Apply AM-GM Inequality on \( b^2 \) and \( c^2 \)**:
\[ \frac{b^2 + c^2}{2} \geq \sqrt{b^2 c^2} = bc \]
This implies:
\[ b^2 + c^2 \geq 2bc \tag{2} \]

3. **Apply AM-GM Inequality on \( c^2 \) and \( d^2 \)**:
\[ \frac{c^2 + d^2}{2} \geq \sqrt{c^2 d^2} = cd \]
This implies:
\[ c^2 + d^2 \geq 2cd \tag{3} \]

4. **Apply AM-GM Inequality on \( d^2 \) and \( a^2 \)**:
\[ \frac{d^2 + a^2}{2} \geq \sqrt{d^2 a^2} = da \]
This implies:
\[ d^2 + a^2 \geq 2da \tag{4} \]

5. **Add all inequalities (1), (2), (3), and (4)**:
\[ (a^2 + b^2) + (b^2 + c^2) + (c^2 + d^2) + (d^2 + a^2) \geq 2(ab + bc + cd + da) \]
The left-hand side simplifies to:
\[ 2(a^2 + b^2 + c^2 + d^2) \]
Therefore, we have:
\[ 2(a^2 + b^2 + c^2 + d^2) \geq 2(ab + bc + cd + da) \]

6. **Substituting the given condition**:
Since \( a^2 + b^2 + c^2 + d^2 = 25 \):
\[ 2 \times 25 \geq 2(ab + bc + cd + da) \]
This simplifies to:
\[ 50 \geq 2(ab + bc + cd + da) \]
Dividing both sides by 2 gives:
\[ 25 \geq ab + bc + cd + da \]

### Conclusion:
Thus, we conclude that:
\[ ab + bc + cd + da \leq 25 \]