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# If A+B+C = ∏ and A,B,C belongs to (0,∏/2) then prove thatsecA +secB +secC≥6

105 Points
10 years ago

Dear manoj,

A,B,C are less than 90 degrees hence

cosA>cosB>cosC (consider A<B<C)

We have, secA<secB<secC

Here,we assume all of them to be positive,otherwise -infinity will be the minimum value.

we have (cosA+cosB+cosC)(secA+secB+secC)>=9   (tschebyhef’s in equality)

We already know the maximum value of cosA+cosB+cosC is 3/2.(we can prove it easily)

=> (secA+secB+secC)>= 9*(2/3)

Hence

(secA+secB+secC)>=6

All the best.

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