if-x-gt-0-then-e-x-e-x-e-x-e-x-2coshx-e-sechx-

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 by mathdanisur last updated on 23/Jul/21