Question Number 188270 by mnjuly1970 last updated on 27/Feb/23
Commented by mnjuly1970 last updated on 27/Feb/23
$$\:\:{in}\:\:{A}\overset{\Delta} {{B}C}\:{prove}\::\: \\ $$$$\:\:\:\frac{\mathrm{1}}{{cos}\left({A}\right)}\:+\frac{\mathrm{1}}{{cos}\left({B}\right)}\:+\frac{\mathrm{1}}{{cos}\left({C}\right)}\:\geqslant\:\mathrm{6} \\ $$$$\: \\ $$$$\:\:\:\:\:{note}\::\:\mathrm{0}\:<\:{A}\:,\:{B}\:,\:{C}\:<\:\mathrm{90}^{°} \\ $$$$ \\ $$
Answered by mahdipoor last updated on 27/Feb/23
$${f}\left({x},{y},{z}\right)=\frac{\mathrm{1}}{{cos}\left({x}\right)}+\frac{\mathrm{1}}{{cos}\left({y}\right)}+\frac{\mathrm{1}}{{cos}\left({z}\right)} \\ $$$${g}\left({x},{y},{z}\right)={x}+{y}+{z}=\mathrm{180}\:\:\:\left({eq}\:{I}\right) \\ $$$${p}=\left({a},{b},{c}\right)\:,\:\:{f}\left({p}\right)={min}\:{f} \\ $$$$\Rightarrow\bigtriangledown\overset{\rightarrow} {{f}}\left({p}\right)=\lambda\bigtriangledown\overset{\rightarrow} {{g}}\left({p}\right) \\ $$$$\Rightarrow\left(−\frac{{sin}\left({a}\right)}{{cos}^{\mathrm{2}} \left({a}\right)}\right){i}+\left(−\frac{{sin}\left({b}\right)}{{cos}^{\mathrm{2}} \left({b}\right)}\right){j}+\left(−\frac{{sin}\left({c}\right)}{{cos}^{\mathrm{2}} \left({c}\right)}\right){k}= \\ $$$$=\lambda\left({i}+{j}+{k}\right)\:\:\:\:\left({eq}\:{II}\right) \\ $$$$\Rightarrow\Rightarrow{eq}\:{I}\:\:\&\:{II}\:\Rightarrow \\ $$$$\frac{{sina}}{{cos}^{\mathrm{2}} {a}}=\frac{{sinb}}{{cos}^{\mathrm{2}} {b}}=\frac{{sinc}}{{cos}^{\mathrm{2}} {c}}=−\lambda \\ $$$${a}+{b}+{c}=\mathrm{180} \\ $$$$\Rightarrow\Rightarrow{a}={b}={c}=\mathrm{60}\Rightarrow{min}\:{f}=\frac{\mathrm{3}}{{cos}\left(\mathrm{60}\right)}=\mathrm{6} \\ $$