Prove:$\left( 1+\frac{1}{x} \right)^x\left[ x\ln \left( 1+\frac1x \right)+\frac1{1+x} \right]<\mathrm{e},\quad x\in (0,+\infty )$.

Question

Prove inequality: ${{\left( 1+\dfrac{1}{x} \right)}^{x}}\left[ x\ln \left( 1+\dfrac{1}{x} \right)+\dfrac{1}{1+x} \right]<\operatorname{e},\quad x\in (0,+\infty )$.

Answer 1

Let $f(x)={{\left( 1+\dfrac{1}{x} \right)}^{x}}\left( x\ln \left( 1+\dfrac{1}{x} \right)+\dfrac{1}{1+x} \right)$.

Hence, $$f'(x)=\frac{\left(1+\frac{1}{x}\right)^x\left(x(x+1)^2\ln^2\left(1+\frac{1}{x}\right)+2(x+1)\ln\left(1+\frac{1}{x}\right)-x-3\right)}{(x+1)^2}.$$ We'll prove that $$x(x+1)^2\ln^2\left(1+\frac{1}{x}\right)+2(x+1)\ln\left(1+\frac{1}{x}\right)-x-3>0$$ or $$\ln\left(1+\frac{1}{x}\right)>\frac{-x-1+\sqrt{(x+1)^2+x(x+1)^2(x+3)}}{x(x+1)^2}$$ or $$\ln\left(1+\frac{1}{x}\right)>\frac{\sqrt{x^2+3x+1}-1}{x(x+1)}.$$ Now, let $$g(x)=\ln\left(1+\frac{1}{x}\right)-\frac{\sqrt{x^2+3x+1}-1}{x(x+1)}.$$ Thus, $$g'(x)=\frac{2x^3+9x^2+7x-2-2\sqrt{(x^2+3x+1)^3}}{2x^2(x+1)^2\sqrt{x^2+3x+1}}<0$$ because for which we need to prove that $$2\sqrt{(x^2+3x+1)^3}>2x^3+9x^2+7x-2,$$ which is obvious for $2x^3+9x^2+7x-2\leq0$, but for $2x^3+9x^2+7x-2>0$ it gives $$4(x^2+3x+1)^3>(2x^3+9x^2+7x-2)^2$$ or $$x(11x^3+62x^2+107x+64)>0.$$ Id est, $g$ decreases and since $\lim\limits_{x\rightarrow+\infty}g(x)=0$, we got that $f$ increases and since $$\lim_{x\rightarrow+\infty}f(x)=e,$$ we are done!