A proof of the inequality using properties of $\pi$ and $e$, for example, is what I'm looking for. Not calculator approximations showing the inequality holds.

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    $\begingroup$ That is a curious thing to ask. The most natural approach would be to first use known properties of $\pi$ and $e$ to come up with a rough approximation, and then use these approximations to approximate both sides. Note that the numbers are not algebraic, so I wouldn't expect $\pi^4 + \pi^5$ not $e^5$ to be anything nice. $\endgroup$ – Jakub Konieczny Sep 18 '12 at 12:55
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    $\begingroup$ It is interesting that the two quantities are so close. $\endgroup$ – Stephen Herschkorn Jul 3 '13 at 0:37

Machin's formula for $\pi$ is $$ \pi = 16 \arctan\frac{1}{5} - 4 \arctan\frac{1}{239}, $$ where $$ \arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} \pm \cdots. $$ For $x \in (0,1)$, the series for $\arctan$ is a Leibniz series, and so if we stop it an a positive term we get an upper bound, if we stop it at a negative term we get a lower bound. This implies that $$ \pi < \pi_0 = 16 \left(\frac{1}{5} - \frac{1}{3\cdot 5^3} + \frac{1}{5\cdot 5^5} - \frac{1}{7\cdot 5^7} + \frac{1}{9\cdot 5^9}\right) - 4 \left(\frac{1}{239} - \frac{1}{3\cdot 239^3} + \frac{1}{5\cdot 239^5} - \frac{1}{7\cdot 239^7}\right) = 17218914588448662618112448/5480950692586369095703125. $$ On the other hand, clearly $$ e > e_0 = \sum_{k=0}^{13} \frac{1}{k!} = 8463398743/3113510400. $$ Calculation shows that $e_0^6 - \pi_0^5 - \pi_0^4$ is equal to 1018224863065019575410675786199225483509861227579368537136263346296216701037017521692657615552228398812312175603740244153839327500226566100058381471958661659/18596816279019949153023047082167290760065663462355245977908373497676350985555730324462494476619445940383152292789734912000000000000000000000000000000000000000000000, which is a positive fraction. Hence $$ \pi^4+\pi^5 < \pi_0^4 + \pi_0^5 < e_0^6 < e^6. $$


You can do it only by approximations. In this case you can use that $\pi<3.14159267$ and $e>2.71828182$. So $\pi^4+\pi^5<3.14159267^4+3.14159267^5$ and $e^6>2.71828182^6$. Hence it suffices to prove that :$$\left(\frac{314159267}{100000000}\right)^4 + \left(\frac{314159267}{100000000}\right)^5 <\left(\frac{271828182}{100000000}\right)^6$$ After multiplication by $100000000^6$ and dividing by the common factor of 64 this is equivalent to:$$12500000^2*314159267^4 + 1562500*314159267^5 - 135914091^6<0$$

This is as good as it can get. Plug it in some calculation software to get the result...$-1766486976776637078635986938150126848541$. This may seem a large number but relatively speaking it is very small it is about $3*10^{-10}$ from $135914091^6$.


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