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Question Number 147371 by mathdanisur last updated on 20/Jul/21

if  x>0  then:  e^(x+e^(−x) )  + e^(−x+e^x )  ≥ 2coshx ∙ e^(sechx)

$${if}\:\:{x}>\mathrm{0}\:\:{then}: \\ $$ $${e}^{\boldsymbol{{x}}+\boldsymbol{{e}}^{−\boldsymbol{{x}}} } \:+\:{e}^{−\boldsymbol{{x}}+\boldsymbol{{e}}^{\boldsymbol{{x}}} } \:\geqslant\:\mathrm{2}{coshx}\:\centerdot\:{e}^{\boldsymbol{{sechx}}} \\ $$

Answered by mindispower last updated on 20/Jul/21

a+b≥2(√(ab))  ⇒a=e^(x+e^(−x) ) ,b=e^(−x+e^x )   ≥2(√e^(x+e^(−x) −x+e^x ) )=2e^(ch(x))   2e^(ch(x)) ≥2ch(x)e^(1/(ch(x))) ......(?)  t=ch(x)≥ch(0)=1  2e^x ≥2xe^(1/x) ....∀x≥1⇔  e^(x−(1/x)) ≥x,x≥1  f(x)=e^(x−(1/x)) −x  f′(x)=(1+(1/x^2 ))e^(x−(1/x)) −1≥  x≥1⇒x−(1/x)≥1−1=0  ⇒f′(x)≥1+(1/x^2 )−1>0  f is increasing   f(x)≥f(1)=e^0 −1=0  ⇒f(x)≥0 then  just deduce

$${a}+{b}\geqslant\mathrm{2}\sqrt{{ab}} \\ $$ $$\Rightarrow{a}={e}^{{x}+{e}^{−{x}} } ,{b}={e}^{−{x}+{e}^{{x}} } \\ $$ $$\geqslant\mathrm{2}\sqrt{{e}^{{x}+{e}^{−{x}} −{x}+{e}^{{x}} } }=\mathrm{2}{e}^{{ch}\left({x}\right)} \\ $$ $$\mathrm{2}{e}^{{ch}\left({x}\right)} \geqslant\mathrm{2}{ch}\left({x}\right){e}^{\frac{\mathrm{1}}{{ch}\left({x}\right)}} ......\left(?\right) \\ $$ $${t}={ch}\left({x}\right)\geqslant{ch}\left(\mathrm{0}\right)=\mathrm{1} \\ $$ $$\mathrm{2}{e}^{{x}} \geqslant\mathrm{2}{xe}^{\frac{\mathrm{1}}{{x}}} ....\forall{x}\geqslant\mathrm{1}\Leftrightarrow \\ $$ $${e}^{{x}−\frac{\mathrm{1}}{{x}}} \geqslant{x},{x}\geqslant\mathrm{1} \\ $$ $${f}\left({x}\right)={e}^{{x}−\frac{\mathrm{1}}{{x}}} −{x} \\ $$ $${f}'\left({x}\right)=\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){e}^{{x}−\frac{\mathrm{1}}{{x}}} −\mathrm{1}\geqslant \\ $$ $${x}\geqslant\mathrm{1}\Rightarrow{x}−\frac{\mathrm{1}}{{x}}\geqslant\mathrm{1}−\mathrm{1}=\mathrm{0} \\ $$ $$\Rightarrow{f}'\left({x}\right)\geqslant\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{1}>\mathrm{0} \\ $$ $${f}\:{is}\:{increasing}\: \\ $$ $${f}\left({x}\right)\geqslant{f}\left(\mathrm{1}\right)={e}^{\mathrm{0}} −\mathrm{1}=\mathrm{0} \\ $$ $$\Rightarrow{f}\left({x}\right)\geqslant\mathrm{0}\:{then} \\ $$ $${just}\:{deduce} \\ $$ $$ \\ $$ $$ \\ $$

Commented bymathdanisur last updated on 23/Jul/21

please Sir

$${please}\:{Sir} \\ $$

Commented bymathdanisur last updated on 20/Jul/21

thank you Ser, please it′s extent

$${thank}\:{you}\:{Ser},\:{please}\:{it}'{s}\:{extent} \\ $$

Commented bymindispower last updated on 21/Jul/21

just replace x=ch(t)

$${just}\:{replace}\:{x}={ch}\left({t}\right) \\ $$

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