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The inequality goes like this: $$\sqrt7^{\sqrt5}>\sqrt5^{\sqrt7}$$ I have to prove this without using approximate numbers and I just cannot find out how.

Andreas Rejbrand
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ladna
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    See https://math.stackexchange.com/questions/1705690/prove-that-xy-yx – lab bhattacharjee Feb 20 '18 at 11:56
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    @labbhattacharjee While relevant, I suspect that if he's not allowed to use "approximate numbers", then the logarithms of $\sqrt 7$ and $\sqrt5$ are also not available for use. – Arthur Feb 20 '18 at 12:00
  • What the? It's not a duplicate, why would we answer here. And this question has got much better responses. – King Tut Feb 21 '18 at 02:45

4 Answers4

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$$7^5=16807>15625=5^6\\7^5>5^6\\\sqrt7^5=(\sqrt7^{\sqrt 5})^{\sqrt 5}>\sqrt5^6=\sqrt5^{\sqrt {36}}>\sqrt5^{\sqrt {35}}=(\sqrt5^{\sqrt {7}})^{\sqrt{5}}\\\sqrt7^{\sqrt 5}>\sqrt5^{\sqrt 7}$$

Mostafa Ayaz
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This inequality is equivalent to $$\sqrt 5\ln(\sqrt7) > \sqrt 7\ln(\sqrt5)\iff \frac{\ln(\sqrt7)}{\sqrt7}>\frac{\ln(\sqrt 5)}{\sqrt5}.$$

Consider the function $\;f(x)=\dfrac{\ln x}x$. As $\;f'(x)=\dfrac{1-\ln x}{x^2}$, this function is increasing on $(0,\mathrm e]$. Observe that $$\sqrt 5<\sqrt 7<2.7<\mathrm e,\enspace\text{so }\enspace f(\sqrt 5)<f(\sqrt 7). $$

Bernard
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  • This is easier, can you please check my method? – King Tut Feb 20 '18 at 15:10
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    It works fine too and is very close to what I propose. Just a minor remark: the computation of $f'(x)$ seems to use a formula I do not know, and it might hard to follow for some people. Perhaps you should mention it. – Bernard Feb 20 '18 at 15:40
  • Even I don't know it's name, but it goes like differentiating power and base one by one, each time treating other as constant. Frankly taking logarithm and then differentiating would have been much easier :) – King Tut Feb 21 '18 at 02:52
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Consider the function $f(x)=x^{1/x}$. So find derivative and set it zero to get increasing interval. $$f' (x)= \frac{x^{1/x-1}}{x} -\frac{x^{1/x} \ln(x)}{x^2}$$

From here, note that only root of $f'(x)$ is $e$, and $f$ is increasing for $x<e$. Since $e>\sqrt 7 > \sqrt 5$ we can say

$$f(\sqrt7) > f(\sqrt5 ) \\ \implies \sqrt{7}^{1/\sqrt 7} > \sqrt5 ^{1/\sqrt 5}$$

Now raise both sides to power $\sqrt 5 \sqrt 7$ to get the desired result as $$\sqrt 7 ^ \sqrt 5 > \sqrt 5 ^ \sqrt 7$$

King Tut
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write your inequality in the form $$\frac{\sqrt{5}}{\ln(\sqrt{5})}>\frac{\sqrt{7}}{\ln(\sqrt{7})}$$ and consider $$f(x)=\frac{x}{\ln(\sqrt{x})}$$

Dr. Sonnhard Graubner
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