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

Thread starter utkarshakash
Start date Aug 16, 2013
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Inequality

AI Thread Summary

In the discussion, participants focus on proving the inequality cos2A + cos2B + cos2C ≥ -3/2 for triangle ABC using vector methods. A key approach involves transforming the left-hand side into a vector form, specifically (cos2A i + cos2B j + cos2C k) · (i + j + k). The relationship 2A + 2B + 2C = 360° is highlighted as a crucial hint for the proof. Additionally, visualizing the problem by drawing a circle is suggested as a helpful strategy. The conversation emphasizes the application of vector analysis in geometric inequalities.

Aug 16, 2013

utkarshakash

Gold Member

Homework Statement

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

Homework Equations

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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Aug 16, 2013

tiny-tim

Science Advisor

Homework Helper

hi utkarshakash!

hint: 2A + 2B + 2C = 360° …

so draw a circle!

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