Prove $expleft(frac{-x^2}{2c}right) 0$ The 2019 Stack Overflow Developer Survey Results Are...

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Prove $expleft(frac{-x^2}{2c}right) 0$



The 2019 Stack Overflow Developer Survey Results Are InProve inequality: When $n > 2$, $n! < {left(frac{n+2}{sqrt{6}}right)}^n$Limit of $xleft(left(1 + frac{1}{x}right)^x - eright)$ when $xtoinfty$Proving inequality $sqrt{frac{2a}{b+c}}+sqrt{frac{2b}{c+a}}+sqrt{frac{2c}{a+b}} leq sqrt{3 left(frac{a}{b}+frac{b}{c}+frac{c}{a}right)}$Prove that $left |x(t_0) right |exp left (-int_{t_0}^{t} left |A(t_1) right |mathrm{d}t_1 right )le left | x(t) right |$Prove that $ frac12 < 4sin^2left(frac{pi}{14}right) + frac{1}{4cos^2left(frac{pi}{7}right)} < 2 - sqrt{2} $Trying to prove an absolute value inequality $left | asqrt{2} -b right | > frac{1}{2(a+b)}$Prove $exp left( a - sqrt{a^2+x^{-2}} right) < 2 a x^2, quad a>1, x>0$How to prove $n < left(1+frac{1}{sqrt{n}}right)^n$Showing that $text{exp}(left lvert z right rvert/(leftlvert z rightrvert - 1))leq left lvert 1 + z rightrvert$Is $h:xto expleft[-frac{x^2}{2}right]left(expleft[frac{x^4}{24n}right]-1right)$ increasing?












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$begingroup$


Can someone please tell me how to prove the following inequality?



$$expleft(frac{-x^2}{2c}right) < sqrt{c} expleft(frac{-x^2}{3c}right), qquad c > 0, x in mathbb{R}.$$



Thanks.






real-analysis inequality







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$endgroup$












  • $begingroup$
    Taking logarithm of both sides gives $-frac{x^2}{2c} < frac12 log c - frac{x^2}{3c} implies 3clog c+x^2 > 0$, so this can't be proved unless you put restrictions on $c$ and/or $x$.
    $endgroup$
    – Daniel R
    Sep 12 '13 at 10:54
















0












$begingroup$


Can someone please tell me how to prove the following inequality?



$$expleft(frac{-x^2}{2c}right) < sqrt{c} expleft(frac{-x^2}{3c}right), qquad c > 0, x in mathbb{R}.$$



Thanks.






real-analysis inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    Taking logarithm of both sides gives $-frac{x^2}{2c} < frac12 log c - frac{x^2}{3c} implies 3clog c+x^2 > 0$, so this can't be proved unless you put restrictions on $c$ and/or $x$.
    $endgroup$
    – Daniel R
    Sep 12 '13 at 10:54














0












0








0





$begingroup$


Can someone please tell me how to prove the following inequality?



$$expleft(frac{-x^2}{2c}right) < sqrt{c} expleft(frac{-x^2}{3c}right), qquad c > 0, x in mathbb{R}.$$



Thanks.






real-analysis inequality







share|cite|improve this question











$endgroup$




Can someone please tell me how to prove the following inequality?



$$expleft(frac{-x^2}{2c}right) < sqrt{c} expleft(frac{-x^2}{3c}right), qquad c > 0, x in mathbb{R}.$$



Thanks.






real-analysis inequality




real-analysis inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 21 at 20:35









Daniele Tampieri

2,68721022




2,68721022










asked Sep 12 '13 at 8:51









user94464user94464

211




211












  • $begingroup$
    Taking logarithm of both sides gives $-frac{x^2}{2c} < frac12 log c - frac{x^2}{3c} implies 3clog c+x^2 > 0$, so this can't be proved unless you put restrictions on $c$ and/or $x$.
    $endgroup$
    – Daniel R
    Sep 12 '13 at 10:54


















  • $begingroup$
    Taking logarithm of both sides gives $-frac{x^2}{2c} < frac12 log c - frac{x^2}{3c} implies 3clog c+x^2 > 0$, so this can't be proved unless you put restrictions on $c$ and/or $x$.
    $endgroup$
    – Daniel R
    Sep 12 '13 at 10:54
















$begingroup$
Taking logarithm of both sides gives $-frac{x^2}{2c} < frac12 log c - frac{x^2}{3c} implies 3clog c+x^2 > 0$, so this can't be proved unless you put restrictions on $c$ and/or $x$.
$endgroup$
– Daniel R
Sep 12 '13 at 10:54




$begingroup$
Taking logarithm of both sides gives $-frac{x^2}{2c} < frac12 log c - frac{x^2}{3c} implies 3clog c+x^2 > 0$, so this can't be proved unless you put restrictions on $c$ and/or $x$.
$endgroup$
– Daniel R
Sep 12 '13 at 10:54










1 Answer
1






active

oldest

votes


















1












$begingroup$

The inequality doesn't make sense for $cleq 0$, so we assume $c in mathbb{R}^{>0}$. When $x=0$ the inequality is true when $1<sqrt{c}$ (and $c>0$), that is when $c>1$. Now assume $x neq 0$.



We take the right-hand side and divide by the left-hand side (which is non-zero), and observe that $$sqrt{c}frac{expleft(-frac{x^2}{3c}right)}{expleft(-frac{x^2}{2c}right)}=sqrt{c}expleft(frac{x^2}{2c}-frac{x^2}{3c}right)=sqrt{c}expleft(frac{x^2}{6c}right)=expleft(frac{x^2}{6c}+logsqrt{c}right).$$



So the inequality is true provided that $$expleft(frac{x^2}{6c}+logsqrt{c}right)>1,$$ or equivalently $$x^2>-6clogsqrt{c}$$ since $x^2,c in mathbb{R}^{>0}$. (Note that if $x=0$ and $c>1$, then the above equality is also true.)



So the given inequality holds for all $c in mathbb{R}^{>0}$ and $x in mathbb{R}$ provided $x^2>-6clogsqrt{c}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I suspect there might be other constraints. For any $x$ there will be $c$-values for which the inequality holds, and others for which it does not hold. Seems unnatural.
    $endgroup$
    – Rebecca J. Stones
    Sep 12 '13 at 21:42












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

The inequality doesn't make sense for $cleq 0$, so we assume $c in mathbb{R}^{>0}$. When $x=0$ the inequality is true when $1<sqrt{c}$ (and $c>0$), that is when $c>1$. Now assume $x neq 0$.



We take the right-hand side and divide by the left-hand side (which is non-zero), and observe that $$sqrt{c}frac{expleft(-frac{x^2}{3c}right)}{expleft(-frac{x^2}{2c}right)}=sqrt{c}expleft(frac{x^2}{2c}-frac{x^2}{3c}right)=sqrt{c}expleft(frac{x^2}{6c}right)=expleft(frac{x^2}{6c}+logsqrt{c}right).$$



So the inequality is true provided that $$expleft(frac{x^2}{6c}+logsqrt{c}right)>1,$$ or equivalently $$x^2>-6clogsqrt{c}$$ since $x^2,c in mathbb{R}^{>0}$. (Note that if $x=0$ and $c>1$, then the above equality is also true.)



So the given inequality holds for all $c in mathbb{R}^{>0}$ and $x in mathbb{R}$ provided $x^2>-6clogsqrt{c}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I suspect there might be other constraints. For any $x$ there will be $c$-values for which the inequality holds, and others for which it does not hold. Seems unnatural.
    $endgroup$
    – Rebecca J. Stones
    Sep 12 '13 at 21:42
















1












$begingroup$

The inequality doesn't make sense for $cleq 0$, so we assume $c in mathbb{R}^{>0}$. When $x=0$ the inequality is true when $1<sqrt{c}$ (and $c>0$), that is when $c>1$. Now assume $x neq 0$.



We take the right-hand side and divide by the left-hand side (which is non-zero), and observe that $$sqrt{c}frac{expleft(-frac{x^2}{3c}right)}{expleft(-frac{x^2}{2c}right)}=sqrt{c}expleft(frac{x^2}{2c}-frac{x^2}{3c}right)=sqrt{c}expleft(frac{x^2}{6c}right)=expleft(frac{x^2}{6c}+logsqrt{c}right).$$



So the inequality is true provided that $$expleft(frac{x^2}{6c}+logsqrt{c}right)>1,$$ or equivalently $$x^2>-6clogsqrt{c}$$ since $x^2,c in mathbb{R}^{>0}$. (Note that if $x=0$ and $c>1$, then the above equality is also true.)



So the given inequality holds for all $c in mathbb{R}^{>0}$ and $x in mathbb{R}$ provided $x^2>-6clogsqrt{c}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I suspect there might be other constraints. For any $x$ there will be $c$-values for which the inequality holds, and others for which it does not hold. Seems unnatural.
    $endgroup$
    – Rebecca J. Stones
    Sep 12 '13 at 21:42














1












1








1





$begingroup$

The inequality doesn't make sense for $cleq 0$, so we assume $c in mathbb{R}^{>0}$. When $x=0$ the inequality is true when $1<sqrt{c}$ (and $c>0$), that is when $c>1$. Now assume $x neq 0$.



We take the right-hand side and divide by the left-hand side (which is non-zero), and observe that $$sqrt{c}frac{expleft(-frac{x^2}{3c}right)}{expleft(-frac{x^2}{2c}right)}=sqrt{c}expleft(frac{x^2}{2c}-frac{x^2}{3c}right)=sqrt{c}expleft(frac{x^2}{6c}right)=expleft(frac{x^2}{6c}+logsqrt{c}right).$$



So the inequality is true provided that $$expleft(frac{x^2}{6c}+logsqrt{c}right)>1,$$ or equivalently $$x^2>-6clogsqrt{c}$$ since $x^2,c in mathbb{R}^{>0}$. (Note that if $x=0$ and $c>1$, then the above equality is also true.)



So the given inequality holds for all $c in mathbb{R}^{>0}$ and $x in mathbb{R}$ provided $x^2>-6clogsqrt{c}$.






share|cite|improve this answer











$endgroup$



The inequality doesn't make sense for $cleq 0$, so we assume $c in mathbb{R}^{>0}$. When $x=0$ the inequality is true when $1<sqrt{c}$ (and $c>0$), that is when $c>1$. Now assume $x neq 0$.



We take the right-hand side and divide by the left-hand side (which is non-zero), and observe that $$sqrt{c}frac{expleft(-frac{x^2}{3c}right)}{expleft(-frac{x^2}{2c}right)}=sqrt{c}expleft(frac{x^2}{2c}-frac{x^2}{3c}right)=sqrt{c}expleft(frac{x^2}{6c}right)=expleft(frac{x^2}{6c}+logsqrt{c}right).$$



So the inequality is true provided that $$expleft(frac{x^2}{6c}+logsqrt{c}right)>1,$$ or equivalently $$x^2>-6clogsqrt{c}$$ since $x^2,c in mathbb{R}^{>0}$. (Note that if $x=0$ and $c>1$, then the above equality is also true.)



So the given inequality holds for all $c in mathbb{R}^{>0}$ and $x in mathbb{R}$ provided $x^2>-6clogsqrt{c}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 12 '13 at 12:10

























answered Sep 12 '13 at 11:40









Rebecca J. StonesRebecca J. Stones

21.1k22781




21.1k22781












  • $begingroup$
    I suspect there might be other constraints. For any $x$ there will be $c$-values for which the inequality holds, and others for which it does not hold. Seems unnatural.
    $endgroup$
    – Rebecca J. Stones
    Sep 12 '13 at 21:42


















  • $begingroup$
    I suspect there might be other constraints. For any $x$ there will be $c$-values for which the inequality holds, and others for which it does not hold. Seems unnatural.
    $endgroup$
    – Rebecca J. Stones
    Sep 12 '13 at 21:42
















$begingroup$
I suspect there might be other constraints. For any $x$ there will be $c$-values for which the inequality holds, and others for which it does not hold. Seems unnatural.
$endgroup$
– Rebecca J. Stones
Sep 12 '13 at 21:42




$begingroup$
I suspect there might be other constraints. For any $x$ there will be $c$-values for which the inequality holds, and others for which it does not hold. Seems unnatural.
$endgroup$
– Rebecca J. Stones
Sep 12 '13 at 21:42


















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