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Given $$ \begin{eqnarray*} F(x) &=& \displaystyle\int_{0}^{x} \dfrac{1}{\sqrt{1 + t^2}} dt = \log(x + \sqrt{x^2 + 1})\\ G(x) &=& \displaystyle\int_{1}^{x} \dfrac{1}{t} dt = \log(x) \end{eqnarray*} $$ Prove that $F(x) \geq G(x) $ for all $x \geq 1$.

My initial approach is to integrate them, but my teacher said there is a direct way using the Fundamental Theorem of Calculus, and I have no idea how could I approach this problem? My thought is to use the lower sum and upper sum, but that doesn't seem promising. So could anyone share me some ideas?

Edit
Please ignore the $\log(x + \sqrt{x^2 +1})$ part because it was the result after I integrated $F(x)$.

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    $\begingroup$ If the "given" is really given, then what's problem? We clearly have $$\forall\,x>1\,\,,\,\,x+\sqrt {x^2+1}>x\,\,\Longrightarrow \log(x+\sqrt{x^2+1})>\log x...$$ $\endgroup$ – DonAntonio Dec 5 '12 at 4:12
  • $\begingroup$ @DonAntonio: Sorry for the confusion. The $\log(x + \sqrt{x^2 + 1})$ is after I integrated $F(x)$. So I have to show that $F(x) \geq G(x)$ without referring to $\log(x + \sqrt{x^2 + 1})$ and $\log(x)$. $\endgroup$ – Chan Dec 5 '12 at 4:18
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He was likely referring to the Second Fundamental Theorem of Calculus, which states $$\frac{d}{dx}\int_a^xf(t)dt=f(x)$$. You differentiate both sides to see that $$\frac{d}{dx}\int_0^x\frac{dt}{\sqrt{1+t^2}}=\frac{1}{\sqrt{1+x^2}}\geq\frac{d}{dx}\int_1^x\frac{dt}{t}=\frac{1}{t}$$ Then, you will just have to check the case $x=1$, where you can see that $$\int_0^1\frac{dt}{\sqrt{1+t^2}}>\int_1^1\frac{dt}{t}$$ since $\frac{1}{\sqrt{1+x^2}}$ is positive-definite and the right-hand side is obviously $0$.

Then, we have that since $F(1)>G(1)$ and $\frac{dF}{dx}\geq\frac{dG}{dx}$, $$F(x)\geq G(x)$$ for all $x\geq 1$.

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    $\begingroup$ But it's not true that $1/(\sqrt{x^2+1}) \ge 1/x$. That is true if and only if $x\ge \sqrt{x^2+1} > x$ which is false for $x\ge 1$... $\endgroup$ – Potato Dec 5 '12 at 4:53
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Show that $F(1)\geq G(1)$ and that $F'(x)>G'(x)$ if $x>1$. Having those facts in hand, it shouldn't be too tough.

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    $\begingroup$ I don't think your derivative inequality is true. See my comment on the other answer. $\endgroup$ – Potato Dec 5 '12 at 4:55
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Substituting $t=u-1$ in the expression for $F(x)$, we get \begin{eqnarray*} F(x)&=&\int_1^{x+1}\frac{1}{\sqrt{u^2-2(u-1)}}\,du\\ &\geq& \int_1^x\frac{1}{u}\,du\\ &=&G(x) \end{eqnarray*}

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