Proving Inequality: Vector Method for Cos2A+Cos2B+Cos2C in Triangle ABC

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SUMMARY

The forum discussion focuses on proving the inequality cos2A + cos2B + cos2C ≥ -3/2 for triangle ABC using vector methods. The approach involves transforming the left-hand side of the inequality into a vector form, specifically (cos2A i + cos2B j + cos2C k) · (i + j + k). The hint provided emphasizes the relationship 2A + 2B + 2C = 360°, suggesting the use of geometric visualization, such as drawing a circle, to aid in the proof.

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  • Concept of geometric visualization in proofs.
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Homework Statement


In a triangle ABC, prove by vector method cos2A+cos2B+cos2C≥ -3/2.

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The Attempt at a Solution


I can change the LHS of the inequality to the form
(cos2A i + cos2B j + cos2C k).(i+j+k)
 
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hi utkarshakash! :smile:

hint: 2A + 2B + 2C = 360° …

so draw a circle! :wink:
 

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