# If $\frac{x}{y}+\frac{y}{x}\ge2$ and $\sum_{cyc}\frac{x}{y+z}\ge\frac32$, can we say $\sum_{cyc}\frac{w}{x+y+z}\ge\frac43$?

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.

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.

• How did you assume that sum of all variables $(a)=$number of all variables $(n)?$ Aug 14 at 10:11
• In my opinion, that's not how WLOG works Aug 14 at 10:13
• Possibly the OP assumes that variables can be rescaled because the relationship is homogeneous. I am hoping ithe OP adds this step. Aug 14 at 10:46
• @abcdefu Otherwise we replace $x_i$ with $nx_i/a$, and we often do this in homogeneous inequalities. Aug 14 at 11:05
• 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. Aug 15 at 6:05

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?

– User
Aug 14 at 10:52

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.

• +1....an attempt to do it without the help of famous inequalities Aug 14 at 12:33

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

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}

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.

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.

• 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... Aug 14 at 20:30