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Question-174668




Question Number 174668 by Shrinava last updated on 07/Aug/22
Answered by behi834171 last updated on 08/Aug/22
ΣsecA≥3((ΠsecA))^(1/3) ≥3((sec^3 (𝛑/3)))^(1/3) =6  ΣcscA≥3((ΠcscA))^(1/3) ≥3((csc(𝛑/3)))^(1/3) =2(√3)  ⇒LHS≥(√3)×6+9×2(√3)=24(√3)  .■
$$\Sigma\boldsymbol{{secA}}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\Pi\boldsymbol{{secA}}}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\boldsymbol{{sec}}^{\mathrm{3}} \frac{\boldsymbol{\pi}}{\mathrm{3}}}=\mathrm{6} \\ $$$$\Sigma\boldsymbol{{cscA}}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\Pi\boldsymbol{{cscA}}}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\boldsymbol{{csc}}\frac{\boldsymbol{\pi}}{\mathrm{3}}}=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\boldsymbol{{LHS}}\geqslant\sqrt{\mathrm{3}}×\mathrm{6}+\mathrm{9}×\mathrm{2}\sqrt{\mathrm{3}}=\mathrm{24}\sqrt{\mathrm{3}}\:\:.\blacksquare \\ $$
Commented by Shrinava last updated on 08/Aug/22
Cool dear Sir thank you very much
$$\mathrm{Cool}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$

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