Four variables inequality proof

Question

Let $x_1 , x_2 , x_3 , x_4 >0$. $$\text{Given} \quad \sum_{i=1}^{4} \frac{1}{1+x_i} = 3~,\quad \text{prove} \quad \prod_{i=1}^4 x_i \le 3^{-4} .$$

A proof by Jensen's Inequality ?

Answer 1

Let $$f(u) = \frac{1}{1 + \mathrm{e}^u}.$$ We have $f''(u) = \frac{(\mathrm{e}^u - 1)\mathrm{e}^u}{(1 + \mathrm{e}^u)^3} \le 0$ for all $u \le 0$. Thus, $f(u)$ is concave on $u\le 0$.

WLOG, assume that $x_1 \le x_2 \le x_3 \le x_4$.

We split into two cases:

1. If $x_4 \le 1$, using the concavity of $f(u)$, we have $$f(\ln x_1) + f(\ln x_2) + f(\ln x_3) + f(\ln x_4) \le 4\, f\left(\frac{\ln x_1 + \ln x_2 + \ln x_3 + \ln x_4}{4}\right)$$ which results in $$\frac{1}{1 + x_1} + \frac{1}{1 + x_2} + \frac{1}{1 + x_3} + \frac{1}{1 + x_4} \le \frac{4}{1 + \sqrt[4]{x_1x_2x_3x_4}}$$ or $$3 \le \frac{4}{1 + \sqrt[4]{x_1x_2x_3x_4}}$$ or $$x_1x_2x_3x_4 \le 3^{-4}.$$

2. If $x_4 > 1$, using $\frac{1}{1 + x_3} + \frac{1}{1 + x_4} > 1$ or $\frac{1 - x_3 x_4}{(1 + x_3)(1 + x_4)}\ge 0$, we have $x_3 \le 1$. Using the concavity of $f(u)$, we have $$f(\ln x_1) + f(\ln x_2) + f(\ln x_3) \le 3\, f\left(\frac{\ln x_1 + \ln x_2 + \ln x_3}{3}\right)$$ which results in $$\frac{1}{1 + x_1} + \frac{1}{1 + x_2} + \frac{1}{1 + x_3} \le \frac{3}{1 + \sqrt[3]{x_1x_2x_3}}$$ or $$3 - \frac{1}{1 + x_4} \le \frac{3}{1 + \sqrt[3]{x_1x_2x_3}}$$ or $$x_1x_2x_3 \le \left(\frac{1}{2 + 3x_4}\right)^3.$$ Thus, we have $$x_1x_2x_3x_4 \le \left(\frac{1}{2 + 3x_4}\right)^3 x_4 \le \frac{1}{125} < 3^{-4}$$ where we have used $$\frac{1}{125} - \left(\frac{1}{2 + 3x_4}\right)^3 x_4 = \frac{(x_4 - 1)(27x_4^2 + 81x_4 - 8)}{125(2 + 3x_4)^3} \ge 0.$$

We are done.

Answer 2

Partial answer

You can try to use Lagrange multipliers to solve $$ \max \left(f(x_1,x_2,x_3,x_4)=x_1 x_2 x_3 x_4\right) $$

such that, $$ \frac{1}{1+x_1} + \frac{1}{1+x_2}+\frac{1}{1+x_3}+\frac{1}{1+x_4}-3 =0 $$

Once you write down the Lagrangian and try to compute its stationary points, you immediately see that:

1. If $x_i > 1$ for all $i$, they must all have the same value but, using the constraint, that value would be $\frac 13$. So, they can not be all > 1.

2. If $x_i \leq 1$ for all $i$, the only solution is $x_i = \frac 13$, which in fact leads to $f = 3^{-4}$.

Now you need to show that any maximum must be a stationary point of the Lagrangian and that any solutions that do not satisfy 1. or 2. satisfy $f \leq 3^{-4}$.

Answer 3

Let $ x_1 = \frac{a}{1-a} , x_2 = \frac{b}{1-b}, x_3 = \frac{c}{1-c}, x_4 = \frac{d}{1-d}$. (Why must $a, b, c, d$ exist?)
Then, from the condition that $ \sum \frac{1}{1 + x_i} = 3$, we get that $ \sum (1 -a ) = 3, $ or that $ a + b + c + d = 1$.

Observe that since $ x_1 = \frac{a}{b+c+d}$, WTS the desired inequality:

$$ abcd \leq \frac{ a+b+c}{3}\cdot \frac{b+c+d}{3} \cdot \frac{c+d+a}{3} \cdot \frac{a+b+c}{3}.$$

This follows immediately from AM-GM applied to each of the terms on the RHS, hence we are done.

---

Note

• This feels like "magic" to me. Obviously the substitution of $ x_ 1 = \frac{a}{1-a} $ was the key, but why should we have done that?
• The naive substitution of $\frac{1}{ 1+x_i} = y_i $ (like I suggested in the comments) would have yielded $ x_i = \frac{ 1- y_i } { y_i } $ for $ 0 < y_i < 1 $ and $ \sum y_i = 3$, but it's not immediately clear how to show that $ 3^4 \prod(1-y_i) \leq \prod y_i$ or even how to apply the condition. I then did another substitution of $ y_i = 1 - z_i$ with $ \sum z_i = 1$, which yielded the above solution.
• $ x_ 1 \geq 1 \Rightarrow a \geq \frac{1}{2} \Rightarrow b < \frac{1}{2} \rightarrow x_2 \leq 1 $ like what River showed in their proof.
• The function $ f(x) = \ln \frac{x}{1-x} $ has second derivative $ \frac{ 2x - 1 } { (x-1)^2x^2}$, so using a Jensen's approach would still have required splitting into cases of $ a > 1/2$, with an identical result to River where we show $ \frac{ a ( 1-a)^2}{(2+a)^3}$ has a maximum of $ \frac{1}{125}$ on the domain $ [ \frac{1}{2} , 1 )$.