Consider a triangle \(ABC\) with area \(\Delta\). Let \(x = r_1 \sec \left( \dfrac A2 \right),y= r_2 \sec \left( \dfrac B2 \right)\) and \(z = r_3\sec \left( \dfrac C2 \right) \), where \(r_1,r_2\) and \(r_3\) are the radii of the excircles of the triangle \(ABC\).

Find the infimum of the function \(f(x,y,z) = \dfrac{xyz(x+y+z)}{\Delta^2} \).

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TopNewest@Vignesh S I'm getting the answer as

\[3\frac { {(abc)}^{\frac{4}{3}}} {{\Delta}^{2}} \]

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The answer is \(\boxed{16}\). Its just equilateral \(\Delta\).But I need a proof for it.

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Can @Otto Bretscher , @Pi Han Goh , @Satyajit Mohanty , @Sandeep Bhardwaj , or anybody else help me out with this. I'm not very good at inequalities, but I tried AM-GM , but its not helping much.

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