Which one is greater $\left(5/2\right)^{2/5}$ or $\left(7/2\right)^{2/7}$?

Question

I have a question here:

Which of $\left(5/2\right)^{2/5}$ and $\left(7/2\right)^{2/7}$ is greater?

I tried comparison by the function $y=x^{1/x}$ and found the derivative as follows $$\frac{\partial y}{\partial x}=\frac{1}{x}(x^{\frac{1}{x}-1})+x^{1/x}\ln x=x^{1/x}(1/x^2+\ln x)$$

I got stuck here, what to do next?

My teacher told me the answer is $\left(5/2\right)^{2/5}$. I want to know how it is.

Answer 1

Raising both numbers to the power $35/2$ and multiplying both by $2^7$ shows that $$(5/2)^{2/5}>(7/2)^{2/7}\qquad\Leftrightarrow\qquad(5/2)^7>(7/2)^5\qquad\Leftrightarrow\qquad5^7>2^2\cdot7^5.$$ The latter inequality is not hard to check by hand.

Answer 2

It is enough to notice that:

$$ f(x) = \log\left(x^{\frac{1}{x}}\right) = \frac{\log x}{x} \tag{1}$$ fulfills: $$ f'(x) = \frac{1-\log x}{x^2},\qquad f''(x) = \frac{2\log x-3}{x^3} \tag{2}$$ hence $f(x)$ is a concave function over the interval $I=\left[\frac{5}{2},\frac{7}{2}\right]$, increasing from $\frac{5}{2}$ to $e$ and decreasing from $e$ to $\frac{7}{2}$. Since the value of $f'''(x)=\frac{11-6\log x}{x^4}$ on $I$ is positive and quite small, the graph of $f(x)$ over $I$ is essentially the graph of a concave-down parabola with vertex at $x=e$. Since $\frac{5}{2}$ is the endpoint of $I$ closer to $e$,

$$ f\left(\frac{5}{2}\right) > f\left(\frac{7}{2}\right).\tag{3}$$

To prove it in a more formal way, compute the second-order Taylor expansion of $f(x)$ centered at $x=e$ and use Lagrange's remainder: $$f(x) = f(e)+f'(e)(x-e)+\frac{f''(e)}{2}(x-e)^2+\frac{f'''(\xi)}{6}(x-e)^3,\qquad \xi\in I.\tag{4}$$ That gives $f(5/2)>\frac{73}{200}>f(7/2)$ and we are done.

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An alternative to prove $5^7>4\cdot 7^5$ (that is equivalent to the previous claim) is to prove:

$$ \left(\frac{1+\frac{1}{6}}{1-\frac{1}{6}}\right)^6<\frac{35}{4}\tag{5}$$ by exploiting the Taylor series of the function $g(x)=\left(\frac{1+x}{1-x}\right)^6$: $$ g(x) = 1+\sum_{n\geq 1}\frac{4n}{15}(23+20n^2+2n^4)\,x^n \tag{6} $$ and proving the inequality: $$ \forall x\in(0,1),\qquad \left(\frac{1+\frac{x}{6}}{1-\frac{x}{6}}\right)^6\leq 1+2x+2x^2+3x^3\tag{7}$$ from which: $$ \left(\frac{1+\frac{1}{6}}{1-\frac{1}{6}}\right)^6\leq 8.\tag{8}$$

Answer 3

Arithmetic mean point of 5/2 and 7/2 is 3. Let us raise each number to $7 \cdot 5$ (which is a monotonically increasing function) and rewrite the two numbers as $$(3-1/2)^{2\cdot 7} = (3-1/2)^{2\cdot 5 + 4} \hspace{0.2cm}\text{ and } \hspace{0.2cm} (3+1/2)^{2 \cdot 5}$$

Write the difference of these two numbers and factor with conjugate rule $$((3-1/2)^{5+2} + (3+1/2)^5)( (3-1/2)^{5+2}- (3+1/2)^5)$$

First factor is obviously positive as it is sum of 2 positives. Expand the second factor with binomial rules, exponent 5 and first term factored out $(3-1/2)^2 = 25/4$

So we got a factor of $25/4-1 = 21/4 > 5$ left of each even term to "compensate" the negative odd terms in the difference. Now it is an easy exercise to this pair wise compensation will be enough to show that the factor is positive, and thus

$$(5/2)^{2/5} > (7/2)^{2/7}$$