If in any triangle, the area `DeltaABC le(b^(2)+c^(2))/(lambda)`, then the largest possible numerical value of `lambda` is

Options

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

Explanation

To solve the problem, we need to find the largest possible numerical value of \( \lambda \) such that the area \( \Delta \) of triangle \( ABC \) satisfies the inequality:

\[ \Delta \leq \frac{b^2 + c^2}{\lambda} \]

### Step-by-Step Solution:

1. **Express the Area of Triangle**:
The area \( \Delta \) of triangle \( ABC \) can be expressed using the formula:
\[ \Delta = \frac{1}{2}bc \sin A \]

2. **Set Up the Inequality**:
Substitute the expression for the area into the inequality:
\[ \frac{1}{2}bc \sin A \leq \frac{b^2 + c^2}{\lambda} \]

3. **Rearrange the Inequality**:
Multiply both sides by \( \lambda \) and rearrange:
\[ \lambda \cdot \frac{1}{2}bc \sin A \leq b^2 + c^2 \]
This can be rewritten as:
\[ \lambda \cdot bc \sin A \leq 2(b^2 + c^2) \]

4. **Isolate \( \lambda \)**:
Now, isolate \( \lambda \):
\[ \lambda \leq \frac{2(b^2 + c^2)}{bc \sin A} \]

5. **Maximize \( \lambda \)**:
To find the maximum value of \( \lambda \), we need to minimize the term \( \frac{bc \sin A}{b^2 + c^2} \). The maximum value of \( \sin A \) is 1 (when \( A = 90^\circ \)). Thus:
\[ \lambda \leq \frac{2(b^2 + c^2)}{bc} \]

6. **Use the Cauchy-Schwarz Inequality**:
By applying the Cauchy-Schwarz inequality:
\[ (b+c)^2 \leq 2(b^2 + c^2) \]
We can deduce that:
\[ \frac{(b+c)^2}{bc} \geq 4 \implies \frac{b^2 + c^2}{bc} \geq 2 \]

7. **Final Calculation**:
Thus:
\[ \lambda \leq 2 \]
Therefore, the largest possible numerical value of \( \lambda \) is \( 2 \).

### Conclusion:
The largest possible numerical value of \( \lambda \) is \( 2 \).