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

$$ifx>0then:$$ $$e^{\mathit{x}+{\mathit{e}}^{-\mathit{x}}}+e^{-\mathit{x}+{\mathit{e}}^{\mathit{x}}}⩾2coshx\cdot e^{\mathit{s}\mathit{e}\mathit{c}\mathit{h}\mathit{x}}$$

Answered by mindispower last updated on 20/Jul/21

$$a+b⩾2\sqrt{ab}$$ $$\Rightarrow a=e^{x+e^{-x}},b=e^{-x+e^x}$$ $$⩾2\sqrt{e^{x+e^{-x}-x+e^x}}=2e^{ch\left(x\right)}$$ $$2e^{ch\left(x\right)}⩾2ch\left(x\right)e^{\frac{1}{ch\left(x\right)}}......(?)$$ $$t=ch\left(x\right)⩾ch\left(0\right)=1$$ $$2e^x⩾2{xe}^{\frac{1}{x}}....\mathrm{\forall }x⩾1\iff$$ $$e^{x-\frac{1}{x}}⩾x,x⩾1$$ $$f\left(x\right)=e^{x-\frac{1}{x}}-x$$ $$f^{\prime }\left(x\right)=(1+\frac{1}{x^2})e^{x-\frac{1}{x}}-1⩾$$ $$x⩾1\Rightarrow x-\frac{1}{x}⩾1-1=0$$ $$\Rightarrow f^{\prime }\left(x\right)⩾1+\frac{1}{x^2}-1>0$$ $$fisincreasing$$ $$f\left(x\right)⩾f\left(1\right)=e^0-1=0$$ $$\Rightarrow f\left(x\right)⩾0then$$ $$justdeduce$$ $$$$ $$$$

Commented bymathdanisur last updated on 23/Jul/21

$$pleaseSir$$

Commented bymathdanisur last updated on 20/Jul/21

$$thankyouSer,please{it}^{\prime }sextent$$

Commented bymindispower last updated on 21/Jul/21

$$justreplacex=ch\left(t\right)$$

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