In Delta ABC,the minimum value of frac{sum co | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $x = 	an \frac{A}{2}$,$y = 	an \frac{B}{2}$,$z = 	an \frac{C}{2}$.In any $\Delta ABC$,we know that $	an \frac{A}{2} 	an \frac{B}{2} + 	a

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let $x = 	an \frac{A}{2}$,$y = 	an \frac{B}{2}$,$z = 	an \frac{C}{2}$.In any $\Delta ABC$,we know that $	an \frac{A}{2} 	an \frac{B}{2} + 	an \frac{B}{2} 	an \frac{C}{2} + 	an \frac{C}{2} 	an \frac{A}{2} = 1$.The given expression is $E = \frac{\frac{1}{x^2 y^2} + \frac{1}{y^2 z^2} + \frac{1}{z^2 x^2}}{\frac{1}{x^2 y^2 z^2}} = x^2 + y^2 + z^2$.We know that $x^2 + y^2 + z^2 \ge xy + yz + zx$.Also,$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$.Using the identity $xy + yz + zx = 1$,we have $x^2 + y^2 + z^2 \ge 1$.However,for a triangle,$x, y, z > 0$ and $xy + yz + zx = 1$. By Cauchy-Schwarz inequality,$x^2 + y^2 + z^2 \ge xy + yz + zx = 1$.Equality holds when $x = y = z = \frac{1}{\sqrt{3}}$,which corresponds to an equilateral triangle where $A = B = C = 60^\circ$.Thus,the minimum value is $1$.

In $\Delta ABC$,the minimum value of $\frac{\sum \cot^2 \frac{A}{2} \cot^2 \frac{B}{2}}{\prod \cot^2 \frac{A}{2}}$ is

Similar Questions

The equation $\sec^2 \theta = \frac{4xy}{(x + y)^2}$ is only possible when

If $A+B+C=\frac{3 \pi}{2}$,then $\cos 2 A+\cos 2 B+\cos 2 C=$

If $\alpha \in \left( 0, \frac{\pi}{2} \right)$,then $\sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}$ is always greater than or equal to

If $0 < \theta < \pi$,then the minimum value of $3 \sin \theta + \text{cosec}^3 \theta$ is

The minimum value of $2^{\sin x} + 2^{\cos x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module