prove that acosA+bcosB+ccosC<=s

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Please find below the solution to the asked query:

Let us assume thatacosA+bcosB+cosCs is true.If this assumption gives valid result then our assumption will be true.Let 'c' be largest sideacosA+bcosB+cosCs acosA12a ab2+c2-a22bc12aab2+c2-a2abca2b2+c2-a2a2bca2b2+a2c2-a4a2bca4-a2b2-a2c2+a2bc0Put c=b as it is longest sidea4-a2b2-a2b2+a2bc0a4-2a2b2+a2bc-a3c+a3c+a3b-a3b0a4-a3b-a3c+a2bc+ a3b+a3c-2a2b20which is Schur inequality, henceaba-b20 which is true as square quantity is always greater than or equal 0Hence our assumption was correct.acosA+bcosB+cosCs 

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