# Prove that $\frac{\sqrt[n]{\prod_{k = 1}^nx_n}}{m} \ge n - 1$ where $\sum_{k = 1}^n\frac{1}{x_k + m} = \frac{1}{m}$.

Given positives $$x_1, x_2, \cdots, x_{n - 1}, x_n$$ such that $$\large \sum_{k = 1}^n\frac{1}{x_k + m} = \frac{1}{m}$$. Prove that $$\large \frac{\displaystyle \sqrt[n]{\prod_{k = 1}^nx_n}}{m} \ge n - 1$$

We have that $$\sum_{k = 1}^n\frac{1}{x_k + m} = \frac{1}{m} \iff \sum_{k = 1}^n\frac{m}{x_k + m} = 1 \iff \sum_{k = 1}^n\frac{x_k}{x_k + m} = n - 1$$

Let $$x_1 \ge x_2 \ge \cdots \ge x_{n - 1} \ge x_m$$, using the Hölder's inequality, it is seen that $$\left(\sum_{k = 1}^nx_k\right) \cdot \left(\sum_{k = 1}^n\frac{1}{x_k + m}\right) \ge m \cdot \sum_{k = 1}^n\frac{x_k}{x_k + m} \implies \left(\sum_{k = 1}^nx_k\right) \cdot \frac{1}{m} \ge m(n - 1)$$

Unfortunately, I can't go for more.

Let $$y_k = \dfrac{m}{x_k + m}, 0 < y_k <1, \sum{y_k}=1$$. Since $$x_k=m \cdot \left(\dfrac{1-y_k}{y_k}\right)$$, the inequality to be proven translates as $$\prod{(1-y_k)}\ge (n-1)^n\prod{y_k}$$.

Noting that $$1-y_k =\sum {y_q}-y_k$$, applying the mean inequality gives $$(1-y_k)^{n-1} \ge (n-1)^{n-1}\dfrac{\prod {y_q}}{y_k}$$ for eack $$k=\overline{1, n}$$.

Multiplying all these inequalities and taking the root of order $$\dfrac{1}{n-1}$$ at the end we are done!

• (+1) Nice approach. I have added an AM-GM proof that is essentially a fleshing out of your answer. If you think it is inappropriate, I will remove it.
– robjohn
Commented Sep 24, 2019 at 19:13
• @robjohn No need, it's useful adding more detail Commented Sep 24, 2019 at 19:22

$$\sum_{k\neq i}\frac{1}{x_k+m}=\frac{1}{m}-\frac{1}{x_i+m}=\frac{x_i}{m(x_i+m)}.$$ Thus, by AM-GM $$\prod_{i=1}^n\frac{x_i}{x_i+m}=m^n\prod_{i=1}^n\sum_{k\neq i}\frac{1}{x_k+m}\geq\frac{m^n(n-1)^n}{\prod\limits_{i=1}^n\left(\prod\limits_{k\neq i}(x_k+m)\right)^{\frac{1}{n-1}}}=\frac{m^n(n-1)^n}{\prod\limits_{i=1}^n(x_i+m)},$$ which ends a proof.

Lemma: If $$u_k\gt0$$ and $$\sum_{k=1}^n\frac1{1+u_k}=1\tag1$$ then $$\prod_{k=1}^nu_k\ge(n-1)^n\tag2$$

To answer the question: apply the Lemma with $$u_k=x_k/m$$.

$$\boldsymbol{u_k\lt1}$$ Does Not Minimize $$\boldsymbol{(2)}$$:

Condition $$(1)$$ allows only one of the $$u_k$$ to be less than $$1$$. Suppose that $$u_1\lt1$$. Then \begin{align} 0 &\ge\frac1{1+u_1}+\frac1{1+u_2}-1\tag3\\ &=\frac{1-u_1u_2}{(1+u_1)(1+u_2)}\tag4\\ &\ge\frac{1-u_1u_2}{2+2u_1u_2}\tag5\\ &=\frac12+\frac1{1+u_1u_2}-1\tag6 \end{align} Explanation:
$$(3)$$: apply $$(1)$$
$$(4)$$: algebra
$$\phantom{(4)\text{:}}$$ this implies $$u_1u_2\ge1$$
$$(5)$$: $$(1+u_1)(1+u_2)-(2+2u_1u_2)=(1-u_1)(u_2-1)\ge0$$
$$\phantom{(5)\text{:}}$$ and $$1-u_1u_2\le0$$
$$(6)$$: algebra

If $$u_1\lt1$$, let $$u$$ be so that $$\frac12+\frac1{1+u}=\frac1{1+u_1}+\frac1{1+u_2}\le1$$. Then $$u\ge1$$ and $$(6)$$ says that \begin{align} \frac12+\frac1{1+u} &=\frac1{1+u_1}+\frac1{1+u_2}\tag7\\ &\ge\frac12+\frac1{1+u_1u_2}\tag8 \end{align} Thus, we can replace $$u_1$$ and $$u_2$$ by $$1$$ and $$u$$, which maintains $$(1)$$ and reduces $$(2)$$ (because $$(8)$$ says that $$u\le u_1u_2$$). So to minimize $$(2)$$, we can assume $$u_k\ge1$$ for all $$k$$.

Here are three proofs of the Lemma, one using convexity, one using variational methods, and one using AM-GM.

Convexity Proof:

Note that $$(1+x)\log(x)$$ is convex for $$x\ge1$$. Therefore, since we can assume $$u_k\ge1$$, \begin{align} \sum_{k=1}^n\log(u_k) &=\sum_{k=1}^n(1+u_k)\log(u_k)\frac1{1+u_k}\tag9\\ &\ge\left(1+\sum_{k=1}^n\frac{u_k}{1+u_k}\right)\log\left(\sum_{k=1}^n\frac{u_k}{1+u_k}\right)\tag{10}\\[6pt] &=n\log(n-1)\tag{11} \end{align} Explanation:
$$\phantom{1}(9)$$: multiply and divide each term by $$1+u_k$$
$$(10)$$: apply Jensen's Inequality to $$(1+x)\log(x)$$
$$(11)$$: use $$(1)$$ and $$\frac{u_k}{1+u_k}=1-\frac1{1+u_k}$$

$$\large\square$$

Variational Proof: Restricted by $$(1)$$, the variations of $$u_k$$ must satisfy $$\sum_{k=1}^n\frac{\delta u_k}{(1+u_k)^2}=0\tag{12}$$ To minimize the product in $$(2)$$, the variations of $$u_k$$ must satisfy $$\sum_{k=1}^n\frac{\delta u_k}{u_k}=0\tag{13}$$ So that every variation that satisfies $$(1)$$ also satisfies $$(2)$$, variational orthogonality requires that there be a constant $$\lambda$$ so that $$(1+u_k)^2=\lambda u_k\tag{14}$$ For any $$\lambda\gt4$$, there are two roots of $$(14)$$. Since the product of these roots is $$1$$, either both roots are $$1$$ or one root must be less than $$1$$ and the other must be greater. As mentioned earlier, we are only concerned with $$u_k\ge1$$. That forces all the $$u_k$$ to be the same. Therefore, $$(1)$$ says that $$u_k=n-1\tag{15}$$ which gives $$(2)$$.

$$\large\square$$

Let $$v_k=\frac1{1+u_k}$$, then $$u_k=\frac{1-v_k}{v_k}$$. $$(1)$$ becomes $$\sum_{k=1}^nv_k=1\tag{16}$$ Furthermore, the AM-GM shows that \begin{align} \frac{1-v_m}{n-1} &=\frac1{n-1}\sum_{k\ne m}v_k\\ &\ge\left(\prod_{k\ne m}v_k\right)^{\frac1{n-1}}\\ &=\left(\frac1{v_m}\prod_{k=1}^nv_k\right)^{\frac1{n-1}}\tag{17} \end{align} Taking the product of $$(18)$$ over $$m$$ from $$1$$ to $$n$$ yields $$\frac1{(n-1)^n}\prod_{m=1}^n(1-v_m)\ge\left(\prod_{m=1}^nv_m\right)^{-\frac1{n-1}}\left(\prod_{k=1}^nv_k\right)^{\frac{n}{n-1}}\tag{18}$$ Cancelling and commuting terms in $$(18)$$ gives \begin{align} \prod_{k=1}^nu_k &=\prod_{m=1}^n\frac{1-v_m}{v_m}\\[6pt] &\ge(n-1)^n\tag{19} \end{align}
$$\large\square$$