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Question

If A, B, C are angles of an acute angle triangle. Then minimum value of
∣ ∣ ∣(tan B+tan C)2tan2Atan2Atan2B(tan C+tan A)2tan2Btan2Ctan2C(tan A+tan B)2∣ ∣ ∣ is

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Solution

Let tan A=a,tanB =b,tanC =c∣ ∣ ∣(b+c)2a2a2b2(c+a)2b2c2c2(a+b)2∣ ∣ ∣
C2C2C1,C3C3C1
=(a+b+c)2∣ ∣ ∣(b+c)2abcabcb2c+ab0c20a+bc∣ ∣ ∣
R1R1(R2+R3)
=(a+b+c)2∣ ∣ ∣2bc2c2bb2c+ab0c20a+bc∣ ∣ ∣
C2C2+1bC1,C3C3+1cC1
=(a+b+c)2∣ ∣ ∣ ∣2bc00b2c+ab2cc2c2ba+b∣ ∣ ∣ ∣=(a+b+c)2×2bc[(a+c)(a+b)bc]
=2abc(a+b+c)3
For ΔABC, we know tan A+tan B+tan C=tan A tan B tan Ca+b+c=abc
Given expression=2(abc)4
The product of three non-zero numbers a, b, c is minimum, when all are equal.
tan A=tanB=tanC=3
Minimum value of the given expression is(=2(3×3×3)4=2×34×32=2×729=1458

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