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$The least value of \sec A + \sec B + \sec C in a acute angled triangle$

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Correct option is D

In ΔABC ,A,B,C are acute. ⇒secA,secB,secC are positive Now, A·M≥H·M.secA+secB+secC3≥3cosA+cosB+cosC →1We have cosA+cosB+cosC3≤cosA+B+C3 ≤cos60°∴ cosA+cosB+cosC≤32 →2From (1) and (2) secA+secB+secC≥6

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The least value of secA+secB+secC in a acute angled triangle

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$The least value of \sec A + \sec B + \sec C in a acute angled triangle$