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Δημοσιεύτηκε: **Τετ Απρ 08, 2026 4:44 am**

Να αποδείξετε ότι

$\displaystyle \frac{1}{2}\left ( e\displaystyle^{\int\limits_{0}^{x}\ln^2\left ( y+\sqrt{1+y^2} \right )dy}-1 \right )\ln \left | \ln\left ( x+\sqrt{1+x^2} \right )\right |>$ $\displaystyle >\left ( 1-x \right )\displaystyle e^{ \int\limits_{0}^{x}\ln^2\left ( y+\sqrt{1+y^2} \right )dy}+\int\limits_{0}^{x}t\ln^2\left ( t+\sqrt{1+t^2} \right )\left ( e^{\int \limits_{0}^{x}\ln^2\left ( y+\sqrt{1+y^2} \right )dy} \right )dt-1,$

για κάθε $\displaystyle x\in \left ( 0,\frac{e^2-1}{2e} \right ].$

Δημοσιεύτηκε: **Τετ Απρ 29, 2026 6:08 pm**

Βάζω τη λύση μου
Έστω $f(x)=\sin\mathrm{h}\h^{-1}(x)$ και $\displaystyle p\left ( x \right )=\left ( e^{\int \limits_{0}^{x}f^2\left ( y \right )dy}-1 \right )\ln \left | f\left ( x \right )\right |,\forall x\in \mathbb{R}_{> 0}\overset{f \pi \alpha \varrho \alpha \gamma \omega \gamma \iota \sigma \iota \mu \eta \mathbb{R}_{> 0}}\Rightarrow \frac{dp\left ( x \right )}{dx}=e^{\int \limits_{0}^{x}f^2\left ( y \right )dy}\left ( \left ( 1-e^{-\int \limits_{0}^{x}f^2(y)dy} \right )\frac{f'\left ( x \right )}{f\left ( x \right )}+\ln \left ( f\left ( x \right ) \right )f^2\left ( x \right ) \right )$ $,\forall x\in \mathbb{R}_{> 0},$ όπου $\displaystyle g\left ( x \right )=\left ( 1-e^{-\int\limits_{0}^{x}f^2\left ( y \right )dy} \right )\frac{f'\left ( x \right )}{f\left ( x \right )}+\ln \left ( f\left ( x \right ) \right )f^2\left ( x \right ),\forall x\in \mathbb{R}_{> 0}\Rightarrow$ $\displaystyle \frac{dg\left ( x \right )}{dx}=f\left ( x \right )f'\left ( x \right )\left ( e^{-\int\limits_{0}^{x}f^2\left ( y \right )dy}+\left ( 1-e^{-\int\limits_{0}^{x}f^2(q)dq} \right )\cdot \frac{f(x)f''\left ( x \right )-\left ( f'\left ( x \right ) \right )^{2}}{f^3\left ( x \right )f'\left ( x \right )}+1+2\ln \left ( f\left ( x \right ) \right ) \right ),$ $\forall x\in \mathbb{R}_{> 0}\left ( \ast \right ),$ κι αφού f,f'>0,f''<0

στο $(0,+\infty),$ άρα $\displaystyle \frac{f\left ( x \right )f''\left ( x \right )-\left ( f'\left ( x \right ) \right )^{2}}{f^{3}\left ( x \right )f'\left ( x \right )}< 0\overset{\int \limits_{0}^{x}f^2> 0,\forall x\in \left ( 0,+\infty \right )}\Rightarrow \left ( 1-e^{-\int \limits_{0}^{x}f^2\left ( q \right )dq} \right )\cdot \frac{f\left ( x \right )f''\left ( x \right )-\left ( f'\left ( x \right ) \right )^{2}}{f^{3}\left ( x \right )f'\left ( x \right )}< 0\Rightarrow$ $\displaystyle e^{-\int \limits_{0}^{x}f^2(y)dy}+\left ( 1-e^{-\int \limits_{0}^{x}f^2(q)dq} \right )\cdot \frac{f\left ( x \right )f''\left ( x \right )-\left ( f'\left ( x \right ) \right )^{2}}{f^{3}\left ( x \right )f'\left ( x \right )}+1+2\ln \left ( f\left ( x \right ) \right )< 2\left ( 1+\ln \left ( f\left ( x \right ) \right ) \right ),$ κι επειδή

γνησίως αύξουσα στο $(0,+\infty)$ με $\displaystyle f\left ( \frac{e^{2}-1}{2e} \right )=1\overset{f'> 0\mathbb{R}_{> 0}}\Rightarrow f\left ( x \right )\leq 1,\forall x\in \left ( 0,\frac{e^{2}-1}{2e} \right ]\Rightarrow 2\ln \left ( f\left ( x \right ) \right )\leq 2\left ( f\left ( x \right )-1 \right )\leq 0$ $\displaystyle \leq 2-2f\left ( x \right )< 2\Rightarrow 2+2\ln \left ( f\left ( x \right ) \right )< 4\overset{ff'> 0_{\left ( \ast \right )}}\Rightarrow \frac{d\left ( g\left ( x \right )-2f^2\left ( x \right ) \right )}{dx}< 0,\forall x\in \left ( 0,\frac{e^{2}-1}{2e} \right ]\Rightarrow g\left ( x \right )-2f^2\left ( x \right )\geq g\left ( \frac{e^{2}-1}{2e} \right )-2f^{2}\left ( \frac{e^{2}-1}{2e} \right )=g\left ( \frac{e^{2}-1}{2e} \right )-2> -2\Rightarrow e^{\int \limits_{0}^{x}f^2(y)dy}g\left ( x \right )>$ $\displaystyle e^{\int \limits_{0}^{x}f^2\left ( z \right )dz}\left ( 2f^2(x)-2 \right ),\forall x\in \left ( 0,\frac{e^{2}-1}{2e} \right ]\Leftrightarrow \frac{dp\left ( x \right )}{dx}> 2\left ( e^{\int \limits_{0}^{x}f^2(y)dy}\frac{d}{dx}\left ( \int \limits_{0}^{x}f^2(u)du \right )-e^{\int \limits_{0}^{x}f^2\left ( y \right )dy} \right )=$ $\displaystyle 2\frac{d}{dx}\left ( e^{\int \limits_{0}^{x}f^2\left ( y \right )dy}-\int \left ( e^{\int \limits_{0}^{x}f^2\left ( y \right )dy} \right )\cdot \left ( x \right )'dx \right )=2\frac{d}{dx}\left ( e^{\int \limits_{0}^{x}f^2(y)dy}-xe^{\int \limits_{0}^{x}f^2(t)dt}+\int xd\left ( e^{\int \limits_{0}^{x}f^2(y)dy} \right ) \right )$ $\displaystyle =2\frac{d}{dx}\left ( \left ( 1-x \right )e^{\int \limits_{0}^{x}f^2(y)dy}+\int \limits_{0}^{x}tf^2(t)\left ( e^{\int \limits_{0}^{t}f^2(y)dy} \right )dt \right ),\forall x\in \left ( 0,\frac{e^{2}-1}{2e} \right ]\left ( 1 \right ).$ Είναι $\displaystyle \frac{d}{dx}\left ( p\left ( x \right )-2\left ( \left ( 1-x \right )e^{\int \limits_{0}^{x}f^2(y)dy}+\int \limits_{0}^{x}tf^2(t)\left ( e^{\int \limits_{0}^{t}f^2(y)dy} \right )dt \right ) \right )\overset{(1)}> 0,\forall x\in \left ( 0,\frac{e^{2}-1}{2e} \right ]$ $\displaystyle \Rightarrow p\left ( x \right )-2\left ( \left ( 1-x \right )e^{\int \limits_{0}^{x}f^2(y)dy}+\int \limits_{0}^{x}tf^2\left ( t \right )\left ( e^{\int \limits_{0}^{t}f^2\left ( y \right )dy} \right )dt \right )> \displaystyle \lim_{x \to 0^{+}}\left ( p\left ( x \right )-2\left ( \left ( 1-x \right )e^{\int \limits_{0}^{x}f^2(y)dy}+\int \limits_{0}^{x}tf^2\left ( t \right )\left ( e^{\int \limits _{0}^{t}f^2(p)dp} \right ) \right )dt \right )$ (2),

και $\displaystyle \displaystyle \lim_{x \to 0^{+}}p\left ( x \right )=\displaystyle \lim_{x \to 0^{+}}\left ( \frac{e^{\int \limits_{0}^{x}f^2(y)dy}-1}{x}\cdot x\ln \left ( f\left ( x \right ) \right ) \right ),$ με $\displaystyle \displaystyle \lim_{x \to 0^{+}}\frac{e^{\int \limits_{0}^{x}f^{2}}-1}{x}\overset{\left ( e^{\int \limits_{0}^{x}f^{2}} \right )'=f^{2}\left ( x \right )e^{\int \limits_{0}^{x}f^{2}}}=f^{2}\left ( 0 \right )\cdot e^{0}=0,$ και $\displaystyle \displaystyle \lim_{x \to 0^{+}}\left ( x\ln \left ( f\left ( x \right ) \right ) \right )=\displaystyle \lim_{x \to 0^{+}}\frac{\ln \left ( f\left ( x \right ) \right )}{\frac{1}{x}}\overset{\left ( \frac{-\infty}{+\infty}DHL \right )}=\displaystyle \lim_{x \to 0^{+}}\left ( -\frac{x^2f'\left ( x \right )}{f\left ( x \right )} \right )\overset{\left ( \frac{0}{0}DHL \right )}=$ $\displaystyle \lim_{x \to 0^{+}}\left ( -\frac{2xf'\left ( x \right )+x^2f''\left ( x \right )}{f'\left ( x \right )} \right )\overset{f'(x)=\frac{1}{\sqrt{x^2+1}}}=0,$ συνεπώς $\displaystyle \displaystyle \lim_{x \to 0^{+}}p\left ( x \right )=0\overset{(2)}\Rightarrow \left ( e^{\int \limits_{0}^{x}f^2(y)dy}-1 \right )\ln \left | f\left ( x \right )\right |-2\left ( \left ( 1-x \right )e^{\int \limits_{0}^{x}f^2(s)ds}+\int \limits_{0}^{x}tf^2(t)\left ( e^{\int \limits_{0}^{t}f^2(w)dw} \right ) dt\right )$ $\displaystyle > -2,\forall x\in \left ( 0,\frac{e^{2}-1}{2e} \right ]\Rightarrow$ $\displaystyle \frac{1}{2}\left ( e^{\int \limits_{0}^{x}\ln ^2\left ( y+\sqrt{y^2+1} \right )dy}-1 \right )\ln \left | \ln \left ( x+\sqrt{x^2+1} \right )\right |$ $\displaystyle > \left ( 1-x \right )e^{\int \limits_{0}^{x}\ln^2(y+\sqrt{y^2+1})dy}+\int \limits_{0}^{x}t\ln^2\left ( t+\sqrt{t^2+1} \right )\left ( e^{\int \limits_{0}^{t}\ln^2(y+\sqrt{y^2+1})dy} \right )dt-1,\displaystyle \forall x\in \left ( 0,\frac{e^{2}-1}{2e} \right ].$

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