5
$\begingroup$

We know that $$\frac{x}{y}+\frac{y}{x}\ge2$$ and $$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}\ge\frac32$$ Can we say that $$\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{x+w+z}+\frac{z}{x+y+w}\ge\frac43$$ Or in general, where there are $n$ variables can we say that their sum (in the same format as above) is no less than $\frac{n}{n-1}$$?$

All the variables used here are positive real numbers.

My thoughts:

I think that we might need induction here, and then we can try completing the squares to find the minimum value. Any help is greatly appreciated.

$\endgroup$

6 Answers 6

5
$\begingroup$

Proposition: $x_1,x_2,\dots, x_n\ge0$, then

$$\sum_{i=1}^{n}{\frac{x_i}{a-x_i}} \ge \frac{n}{n-1},$$ where $a=\sum_{i=1}^n x_i.$

Proof: Without loss of generality, we suppose $a=n$(otherwise we replace $x_i$ with $nx_i/a$, and we often do this in homogeneous inequalities), then it suffice to prove $$\begin{aligned} &\sum_{i=1}^{n}{\frac{x_i}{n-x_i}} \ge \frac{n}{n-1}\\ \iff &\sum_{i=1}^{n}{\Big(\frac{n}{n-x_i}-1\Big)} \ge \frac{n}{n-1} \\ \iff &\sum_{i=1}^{n}{\frac{n}{n-x_i}} \ge \frac{n}{n-1}+n=\frac{n^2}{n-1}\\ \iff &\sum_{i=1}^{n}{\frac{1}{n-x_i}} \ge \frac{n}{n-1}\\ \end{aligned}$$ By Cauchy-Schwarz Inequality, we have $$\Big(\sum_{i=1}^{n}{\frac{1}{n-x_i}}\Big)\Big(\sum_{i=1}^n(n-x_i)\Big)\ge n^2,$$ where $$\sum_{i=1}^n(n-x_i)=n^2-n, $$ Therefore $$\sum_{i=1}^{n}{\frac{1}{n-x_i}}\ge\frac{n^2}{n^2-n}=\frac{n}{n-1}.$$ Now we are done with the proof.

$\endgroup$
10
  • $\begingroup$ How did you assume that sum of all variables $(a)=$number of all variables $(n)?$ $\endgroup$
    – abcdefu
    Aug 14 at 10:11
  • $\begingroup$ In my opinion, that's not how WLOG works $\endgroup$
    – abcdefu
    Aug 14 at 10:13
  • 1
    $\begingroup$ Possibly the OP assumes that variables can be rescaled because the relationship is homogeneous. I am hoping ithe OP adds this step. $\endgroup$ Aug 14 at 10:46
  • $\begingroup$ @abcdefu Otherwise we replace $x_i$ with $nx_i/a$, and we often do this in homogeneous inequalities. $\endgroup$
    – Briar Jam
    Aug 14 at 11:05
  • 1
    $\begingroup$ The initial form of Cauchy-Schwarz inequality is: $\big(\sum a_i^2\big)\big(\sum b_i^2 \big) \ge (\sum a_ib_i)^2$. We let $a_i^2=1/(n-x_i)$ and $b_i^2=n-x_i$, then we get the above inequality. $\endgroup$
    – Briar Jam
    Aug 15 at 6:05
2
$\begingroup$

Applying Cauchy-Schwars inequality, we have

$$\left(\overbrace{\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}}^A\right)\left(x(y+z)+y(x+z)+z(y+x)\right)\ge(x+y+z)^2\implies A≥\frac{(x+y+z)^2}{2(xy+yz+xz)}$$

Then let's apply AM-GM inequality:

$$ \begin{cases}x^2+y^2\ge 2xy\\x^2+z^2\ge 2xz\\ y^2+z^2\ge 2yz\end{cases} $$

\begin{align}&2(x^2+y^2+z^2)\ge2(xy+yz+xz)\\ \implies &x^2+y^2+z^2 \ge xy+yz+xz\\ \implies &(x+y+z)^2\ge 3(xy+yz+xz)\end{align}

Hence, we obtain:

$$A≥\frac{(x+y+z)^2}{2(xy+yz+xz)}≥\frac{3(xy+yz+xz)}{2(xy+yz+xz)}=\frac 32$$


Now, let's apply Cauchy-Schwarz again:

$$\left(\overbrace{\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{x+w+z}+\frac{z}{x+y+w}}^A\right)\left(w(x+y+z)+x(w+y+z)+y(x+w+z)+z(x+y+w)\right)\implies A\ge \frac{(x+y+z+w)^2}{2(xy+xz+xw+yz+yw+zw)}$$

Let's apply AM-GM inequality again:

$$ \begin{cases}x^2+y^2\ge 2xy\\x^2+z^2\ge 2xz\\x^2+w^2\ge 2xw\\ y^2+z^2\ge2yz\\y^2+w^2\ge 2yw\\z^2+w^2\ge 2zw\end{cases} $$

\begin{align}&3(x^2+y^2+z^2+w^2)\ge 2(xy+xz+xw+yz+yw+zw)\\ \implies &x^2+y^2+z^2+w^2 \ge \frac 23 (xy+xz+xw+yz+yw+zw)\\ \implies &(x+y+z+w)^2\ge \frac 83 (xy+xz+xw+yz+yw+zw)\end{align}

Finally, we have

$$A\ge \frac{(x+y+z+w)^2}{2(xy+xz+xw+yz+yw+zw)}\implies A\ge \frac 86=\frac 43$$

Now, can you generalize this method using a simple (just a few steps) combinatorics?

$\endgroup$
1
  • $\begingroup$ Please, feel free to ask questions about the method. $\endgroup$
    – User
    Aug 14 at 10:52
2
$\begingroup$

I'm uncertain if this proof is formally valid, but at least informally:

Consider a sequence of at least two positive real numbers $(a_1, a_2, \cdots a_k)$, and define $S_k = \sum a_i$. Then we claim:

$$A = \sum_{i=1}^{k} \frac{a_i}{S_k - a_i} \ge \frac{k}{k-1} \tag{1}$$

Note that $(1)$ implies that $A$ has its minimum when $a_1=a_2=\cdots=a_k$. In the simplest form, all $a_i=1$, $S_k=k$, and $A=k/(k-1)$.

Now, instead of $1$, WLOG set $a_1=1+z$, where $z$ is any real number $z>-1, z\ne 0$. Then, $S_k=k+z$. The restriction $z>-1$ prevents $a_1$ from becoming negative.

Let's find the value of $A$ with this changed $a_1$. It is:

$$A=\frac{z+1}{k-1} + \frac{k-1}{k+z-1}$$

The $k-1$ in the second fraction comes from $k-1$ copies of $1/(k-1+z)$. Both $k-1$ and $k+z-1$ must be positive (as $k \ge 2$ and $z>-1$).

Now, assume the contrary claim: that this set of $a_i$ gives a smaller total for $A$.

$$\frac{z+1}{k-1} + \frac{k-1}{k+z-1} < \frac{k}{k-1}$$

We can rearrange by

$$ \begin{align} (z+1)(k+z-1) + (k-1)^2 &< k(k+z-1) \\ z^2 +kz+k^2-k &< k^2 +kz-k \\ z^2 &< 0 \end{align} $$

Which is a pretty obvious contradiction. Hence, so long as adding $z$ doesn't make $a_1$ negative, $(1)$ holds.


However, this applies only to a single change from the set of $1$s. I'm not quite certain where to go from here, but I see other (possibly more complete) answers have been posted. Hopefully this partial proof is useful.

$\endgroup$
1
  • 1
    $\begingroup$ +1....an attempt to do it without the help of famous inequalities $\endgroup$
    – abcdefu
    Aug 14 at 12:33
1
$\begingroup$

This answer is a generalization of the method used in the first answer given above.

Problem:

Prove that: $$\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\ge\frac{n}{n-1}$$

Proof:

Using AM-GM inequality we have,

$$\begin{align}&\sum_{1\le i<j\le n} (x_i^2+x_j^2)=(n-1)\sum_{1\le i\le n} x_i^2\ge 2\left(\sum_{1\le i < j \le n}x_ix_j\right)\\ \implies &\sum_{1\le i\le n} x_i^2\ge \frac 2{n-1}\left(\sum_{1\le i < j \le n}x_ix_j\right)\\ \implies &\left(\sum_{1\le i\le n} x_i\right)^2\ge\left( 2+\frac 2{n-1}\right)\left(\sum_{1\le i < j \le n}x_ix_j\right)\\ &\qquad\qquad\quad =\frac{2n}{n-1}\left(\sum_{1\le i < j \le n}x_ix_j\right)\end{align}$$

Finally, applying Cauchy-Schwars, we obtain

$$\begin{align}\left(\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\right)\left(\sum_{1\le j\le n}^n \left({x_j}{\sum_{1\le i\le n} x_i-x_j}\right)\right)&\ge \frac{ \left(\sum_{1\le j\le n} x_i\right)^2}{2\left(\sum_{1\le i < j \le n}x_ix_j\right)}\\ &\ge \frac {\frac{2n}{n-1}\left(\sum_{1\le i < j \le n}x_ix_j\right)}{2\left(\sum_{1\le i < j \le n}x_ix_j\right)}\\ &=\boxed {\frac{n}{n-1}.}\end{align}$$

$\endgroup$
1
$\begingroup$

Problem: Let $x_i \ge 0, i= 1, 2, \cdots, n$. Prove that $$\sum_{i=1}^n \frac{x_i}{S - x_i} \ge \frac{n}{n - 1}$$ where $S = \sum_{j=1}^n x_j$.

Proof.

Since the desired inequality is homogeneous, assume that $\sum_{j=1}^n x_j = 1$. We need to prove that $$\sum_{i=1}^n \frac{x_i}{1 - x_i} \ge \frac{n}{n - 1}.$$

We have $$\frac{x_i}{1 - x_i} - \frac{1}{n - 1} - \frac{n^2}{(n-1)^2}\left(x_i - \frac{1}{n}\right) = \frac{(1-nx_i)^2}{(n-1)^2(1 - x_i)} \ge 0$$ which results in $$\sum_{i=1}^n \frac{x_i}{1 - x_i} - \frac{n}{n-1} \ge \frac{n^2}{(n-1)^2}\sum_{i=1}^n\left(x_i - \frac{1}{n}\right) = 0.$$

We are done.

$\endgroup$
0
$\begingroup$

Here's a convex analysis approach. Let $x \in \mathbb{R}^d$ such that $x_i \ge 0$ and $\sum x_i > 0.$ We shall show (like the other answers) that $$ \sum_i \frac{x_i}{ \left(\sum_j x_j\right) -x_i} \ge \frac{n}{n-1}.$$


Without loss of generality (due to the homogeneity of $\frac{x_i}{\sum x_j - x_i}$ ), assume that $\sum x_i = 1$ . Then notice that $x$ is a vector in the simplex on $\mathbb{R}^d,$ i.e. it is a probability vector. We want to analyse the function $$ f(x) := \sum_i \frac{x_i}{1 - x_i}.$$

Consider a random variable $Z$ which takes the value $x_i$ with probability $x_i$. Then $f(x) = \mathbb{E}[\frac{1}{1-Z}]$. Observe that $ u \mapsto \frac{1}{1-u}$ is convex over $[0,1]$. By Jensen's inequality, we conclude that $$ f(x) \ge \frac{1}{1 - \mathbb{E}[Z]} = \frac{1}{1 - \sum x_i^2}.$$ It is easy to show that $\min \sum x_i^2$ over the simplex in $\mathbb{R}^n$ is $1/n$. Immediately we conclude that $$ \min_x f(x) \ge \frac{1}{1 - 1/n} = \frac{n}{n-1}.$$ Of course, this is tight by taking all the $x_i$s to be equal.

$\endgroup$
1
  • $\begingroup$ Oops, I just noticed the precalculus tag on the question. I'll leave this up just because it's a neat argument, but this probably doesn't help OP all that much... $\endgroup$ Aug 14 at 20:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.