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# Prove That : acosA+bcosB+ccosC<=(a+b+c)/2

9 years ago

Using Sine Rule

a/Sin A = b/SinB = c/sin C

So the above question can be written as

aCosA + a (SinBCosB)/SinA + c (SinCCosA)/Sin A

Now using AM- GM inequality (However this can be valid only in the case of acute angled triangels it will not be valid for obtuse angled or right angled triangles )

aCosA + a (SinBCosB)/SinA + c (SinCCosA)/Sin A >= root(a3(CosA CosB Cos C)*(Sin B Sin C)/Sin2A)

withe the equality existing only when the terms r equal

Then on  equality aCosA = a (SinBCosB)/SinA = c (SinCCosA)/Sin A

= aSinACosA= aSinBCosB = aCos C SinC

= Sin ACosA = SinBCos B = Sin c Cos C

= A= B = C =60

aCosA + a (SinBCosB)/SinA + c (SinCCosA)/Sin A >= a/2+b/2+c/2

=> a CosA + B Cos B + Ccos C>= (a+b+c)/2.

The above identity proved s opposite of wht is stated by u.