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

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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

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