Understanding the Problem: Least Value of Secants in an Acute Triangle

The question asks for the minimum value of the sum of the secants of the angles (sec A + sec B + sec C) in an acute angled triangle ΔABC. An acute angled triangle is one where all three interior angles (A, B, and C) are less than 90 degrees or $\frac{\pi}{2}$ radians. We know that the sum of the angles in any triangle is 180 degrees or $\pi$ radians, so A + B + C = $\pi$.

Applying Jensen's Inequality for Convex Functions

To find the least value, we can analyze the properties of the secant function, specifically $f(x) = \sec x$. In the interval $(0, \frac{\pi}{2})$, which corresponds to the angles of an acute triangle, the secant function is convex. This means its second derivative is positive within this range.

Jensen's inequality states that for a convex function $f(x)$ and points $x_1, x_2, ..., x_n$, the following holds:

$$ \frac{f(x_1) + f(x_2) + ... + f(x_n)}{n} \ge f\left(\frac{x_1 + x_2 + ... + x_n}{n}\right) $$

Applying this to our problem with $f(x) = \sec x$ and the angles A, B, C:

$$ \frac{\sec A + \sec B + \sec C}{3} \ge \sec\left(\frac{A+B+C}{3}\right) $$

Calculating the Minimum Value

We know that for any triangle, A + B + C = $\pi$. Substituting this into the inequality:

$$ \frac{\sec A + \sec B + \sec C}{3} \ge \sec\left(\frac{\pi}{3}\right) $$

The value of $\sec(\frac{\pi}{3})$ (or $\sec(60^\circ)$) is 2.

$$ \frac{\sec A + \sec B + \sec C}{3} \ge 2 $$

Multiplying both sides by 3 gives us the minimum possible value for the sum:

$$ \sec A + \sec B + \sec C \ge 6 $$

Condition for Equality

Jensen's inequality provides the minimum value, and equality holds when all the inputs to the function are equal. In this case, equality occurs when A = B = C.

Since A + B + C = $\pi$, the condition A = B = C implies:

$$ 3A = \pi \implies A = \frac{\pi}{3} $$

So, equality holds when A = B = C = $\frac{\pi}{3}$ (or 60 degrees). This represents an equilateral triangle, which is a specific case of an acute angled triangle.

Let's check the sum for an equilateral triangle:

$$ \sec\left(\frac{\pi}{3}\right) + \sec\left(\frac{\pi}{3}\right) + \sec\left(\frac{\pi}{3}\right) = 2 + 2 + 2 = 6 $$

This confirms that the minimum value is indeed 6.

Conclusion

The least value of $\sec A + \sec B + \sec C$ for an acute angled triangle ΔABC is 6.

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