Question Number 147371 by mathdanisur last updated on 20/Jul/21 | ||
$${if}\:\:{x}>\mathrm{0}\:\:{then}: \\ $$ $${e}^{\boldsymbol{{x}}+\boldsymbol{{e}}^{−\boldsymbol{{x}}} } \:+\:{e}^{−\boldsymbol{{x}}+\boldsymbol{{e}}^{\boldsymbol{{x}}} } \:\geqslant\:\mathrm{2}{coshx}\:\centerdot\:{e}^{\boldsymbol{{sechx}}} \\ $$ | ||
Answered by mindispower last updated on 20/Jul/21 | ||
$${a}+{b}\geqslant\mathrm{2}\sqrt{{ab}} \\ $$ $$\Rightarrow{a}={e}^{{x}+{e}^{−{x}} } ,{b}={e}^{−{x}+{e}^{{x}} } \\ $$ $$\geqslant\mathrm{2}\sqrt{{e}^{{x}+{e}^{−{x}} −{x}+{e}^{{x}} } }=\mathrm{2}{e}^{{ch}\left({x}\right)} \\ $$ $$\mathrm{2}{e}^{{ch}\left({x}\right)} \geqslant\mathrm{2}{ch}\left({x}\right){e}^{\frac{\mathrm{1}}{{ch}\left({x}\right)}} ......\left(?\right) \\ $$ $${t}={ch}\left({x}\right)\geqslant{ch}\left(\mathrm{0}\right)=\mathrm{1} \\ $$ $$\mathrm{2}{e}^{{x}} \geqslant\mathrm{2}{xe}^{\frac{\mathrm{1}}{{x}}} ....\forall{x}\geqslant\mathrm{1}\Leftrightarrow \\ $$ $${e}^{{x}−\frac{\mathrm{1}}{{x}}} \geqslant{x},{x}\geqslant\mathrm{1} \\ $$ $${f}\left({x}\right)={e}^{{x}−\frac{\mathrm{1}}{{x}}} −{x} \\ $$ $${f}'\left({x}\right)=\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){e}^{{x}−\frac{\mathrm{1}}{{x}}} −\mathrm{1}\geqslant \\ $$ $${x}\geqslant\mathrm{1}\Rightarrow{x}−\frac{\mathrm{1}}{{x}}\geqslant\mathrm{1}−\mathrm{1}=\mathrm{0} \\ $$ $$\Rightarrow{f}'\left({x}\right)\geqslant\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{1}>\mathrm{0} \\ $$ $${f}\:{is}\:{increasing}\: \\ $$ $${f}\left({x}\right)\geqslant{f}\left(\mathrm{1}\right)={e}^{\mathrm{0}} −\mathrm{1}=\mathrm{0} \\ $$ $$\Rightarrow{f}\left({x}\right)\geqslant\mathrm{0}\:{then} \\ $$ $${just}\:{deduce} \\ $$ $$ \\ $$ $$ \\ $$ | ||
Commented bymathdanisur last updated on 23/Jul/21 | ||
$${please}\:{Sir} \\ $$ | ||
Commented bymathdanisur last updated on 20/Jul/21 | ||
$${thank}\:{you}\:{Ser},\:{please}\:{it}'{s}\:{extent} \\ $$ | ||
Commented bymindispower last updated on 21/Jul/21 | ||
$${just}\:{replace}\:{x}={ch}\left({t}\right) \\ $$ | ||