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

Question

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.

Answer 1

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!

Answer 2

$$\sum_{k\neq i}\frac{1}{x_k+m}=\frac{1}{m}-\frac{1}{x_i+m}=\frac{x_i}{m(x_i+m)}.$$ 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.

Answer 3

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 $u_k=x_k/m$.

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$\boldsymbol{u_k\lt1}$ 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.

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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 $\frac{u_k}{1+u_k}=1-\frac1{1+u_k}$

$\large\square$

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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 $(2)$.

$\large\square$

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The following is merely a fleshing out of Conrad's answer.

AM-GM Proof:

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 $$ \begin{align} \prod_{k=1}^nu_k &=\prod_{m=1}^n\frac{1-v_m}{v_m}\\[6pt] &\ge(n-1)^n\tag{19} \end{align} $$

$\large\square$