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PostPosted: Sun, 25 Oct 2009 21:36:15 UTC 
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Prove following 3 inequalities:

1) $\forall y>x>0:$ $\left(\frac{y-x}{\ln y - \ln x}\right)^{y-x}\leq\frac{(y/e)^y}{(x/e)^x}\leq\left(\frac{x+y}{2}\right)^{y-x}$

2)$\forall x,y\in\mathbb R, y>x:$ $1\leq\frac{1}{y-x}(e^{\frac{y-x}{2}}-e^{\frac{x-y}{2}})\leq\frac{1}{(y-x)^2}(e^{y-x}+e^{x-y}-2)$

3)$\forall y>x>0:$ $(xy)^{y-x}\leq\left(\frac{(y/e)^y}{(x/e)^x}\right)^2\leq\left(\frac{y^2+xy+x^2}{3}\right)^{y-x}$

I believe all 3 are genuine mine, but if someone thinks he has already seen them elsewhere, please point to the source. I would be interested in having a look into it. Also, it would be nice to see if there are different solutions to these problems.

Good luck and have fun ;)


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PostPosted: Sun, 25 Oct 2009 23:22:23 UTC 
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ex falso quodlibet wrote:
Prove following 3 inequalities:

1) $\forall y>x>0:$ $\left(\frac{y-x}{\ln y - \ln x}\right)^{y-x}\leq\frac{(y/e)^y}{(x/e)^x}\leq\left(\frac{x+y}{2}\right)^{y-x}$

2)$\forall x,y\in\mathbb R, y>x:$ $1\leq\frac{1}{y-x}(e^{\frac{y-x}{2}}-e^{\frac{x-y}{2}})\leq\frac{1}{(y-x)^2}(e^{y-x}+e^{x-y}-2)$

3)$\forall y>x>0:$ $(xy)^{y-x}\leq\left(\frac{(y/e)^y}{(x/e)^x}\right)^2\leq\left(\frac{y^2+xy+x^2}{3}\right)^{y-x}$

I believe all 3 are genuine mine, but if someone thinks he has already seen them elsewhere, please point to the source. I would be interested in having a look into it. Also, it would be nice to see if there are different solutions to these problems.

Good luck and have fun ;)


2)

Spoiler:
Let m={y-x}

Then this reduces to:

$1\le {2\over m}\sinh {m\over 2}\le {2(\cosh m-1)\over m^2}

Applying the half-angle identity

$1\le {2\over m}\sqrt{\cosh {m}-1\over 2}\le {2(\cosh m-1)\over m^2}

now bring the {2\over m} under the square root in the middle term.

$1\le \sqrt{2(\cosh {m}-1)\over m^2}\le {2(\cosh m-1)\over m^2}

This is obviously true if we can get that $1\le \sqrt{2(\cosh {m}-1)\over m^2} for every m>0

since all terms are positive, squaring preserves ordering, so examine the resultant expression:

$1\le {2(\cosh {m}-1)\over m^2}

this is true iff

$1+{m^2\over 2}\le \cosh m

Now the result is obvious, because the \cosh m is an even function, so has all even terms in its taylor series, and the second order approximation is 1+{m^2\over 2}+O(m^3), in particular it is evident that the inequality is true, hence--as each step was reversible, the argument shows the inequality to be true.

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