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ACCESSORY. For those who would like to share their passion for history, American Flag engineering studies additional things that they own..Q:
Inequality involving $t$ or $\log t$
Is it true that
$$\log (x+1)\leq x-1, x\geq 1$$
and
$$x^2+1\leq\log ^2(x+1), x\geq 1.$$
The second one we can think as $\sqrt{x+1}\leq \log(x+1)$, but the first one seems strange to me.
A:
we can easily check that
$$ \log(x+1)-x = \log \left(\frac{x}{x+1}\right) \leq 1$$
therefore
$$ \log (x+1) \leq x $$
A:
Your first inequality is true, but the second is false.
$$\begin{eqnarray}
x^2+1&\leq&\log^2(x+1)\\
x^2+1&\leq&(x+1)\log(x+1)\\
(x^2+1)^{\frac{1}{2}}&\leq&(x+1)^{\frac{1}{2}}\log(x+1)\\
x^{\frac{1}{2}}+1&\leq&(x+1)^{\frac{1}{2}}\\
x+1&\leq&e^{ -1}(x^{\frac{1}{2}}+1)^2\\
(x+1)^2-x^2&\leq&e^{ -1}(x^{\frac{1}{2}}+1)^2\\
(x+1)^2-x^2&\leq&e^{ -1}(2x+1)^2\\
x^2-2x+1&\leq&e^{ -1}(2x+1)^2\\
x^2+1-2x&\leq&e^{ -1}(2x+1)^2\\
x-1&\leq&e^{ -1}
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