Question

In an acute angled triangle abc the least value of seca+secb+secc is

Answer 1

Answer:

Hence, the minimum value is:

6

Step-by-step explanation:

We know that:

$\cos A+\cos B+\cos C\leq \dfrac{3}{2}$

( since by the 5

It is well-known that the sum of the distances of the circumcenter from the sides of a triangle equals the sum of the circumradius(R) and the inradius(r), hence:

$R(\cos A+\cos B+\cos C)=R+r\\\\\cos A+\cos B+\cos C=1+\dfrac{r}{R}$

Also by Euler Theorem we have:

R≥2r

⇒ r/R≤1/2

Hence,

$\cos A+\cos B+\cos C=1+\dfrac{r}{R}\leq 1+\dfrac{1}{2}=\dfrac{3}{2}$

)

Now using the fact that Arithmetic mean is greater than the geometric mean we have:

$\dfrac{\cos A+\cos B+\cos C}{3}\geq (\cos A.\cos B.\cos C)^{\frac{1}{3}}\\\\\dfrac{3}{2}\times \dfrac{1}{3}\geq (\cos A.\cos B.\cos C)^{\frac{1}{3}}\\\\\dfrac{1}{2}\geq (\cos A.\cos B.\cos C)^{\frac{1}{3}}$

This means:

$\cos A.\cos B.\cos C\leq \dfrac{1}{2^3}=\dfrac{1}{8}$

This means that:

$8\leq \dfrac{1}{\cos A.\cos B.\cos C}=\sec A.\sec B.\sec C$

Again on applying Arithmetic Mean is greater than equal to geometric mean we have:

$\dfrac{\sec A+\sec B+\sec C}{3}\geq(\sec A.\sec B.\sec C)^{\frac{1}{3}}\geq8^{\frac{1}{3}}=2\\\\\sec A+\sec B+\sec C\geq6$

Hence, the minimum value is 6.