# If $x_1, \ldots, x_6$ are positive real numbers that add up to $2$. Show that:

If $$x_1,x_2,x_3,x_4,x_5$$ and $$x_6$$ are positive real numbers that add up to $$2$$, then:

$$2^{12} \leq \left(1+\dfrac{1}{x_1}\right) \left(1+\dfrac{1}{x_2}\right)\left(1+\dfrac{1}{x_3}\right)\left(1+\dfrac{1}{x_4}\right)\left(1+\dfrac{1}{x_5}\right)\left(1+\dfrac{1}{x_6}\right) .$$

Factoring out the $$\dfrac{1}{x_1}, \ldots, \dfrac{1}{x_6}$$ from each term, I get $$\dfrac{1}{x_1}(x_1 + 1)\dfrac{1}{x_2}(x_2 + 1)\dfrac{1}{x_3}(x_3 + 1)\dfrac{1}{x_4}(x_4 + 1)\dfrac{1}{x_5}(x_5 + 1)\dfrac{1}{x_6}(x_6 + 1)$$ We know that: $$x_1 + \cdots + x_6 = 2$$

Also, I see that $$x_1, \ldots, x_6$$ are small numbers and $$1$$ over something small will give something big.

How can I keep going?

• maybe the worst case is that each $x_i=1/3$ then it is equal. – derivative Nov 19 '13 at 20:01
• So, what are the biggest numbers that you can choose for the x's? (Hint: it's the same value for all of the x's.) Can you show that this sextuple produces a global minimum given your constraint? – John Nov 19 '13 at 20:02
• Lagrange multipliers, maybe? – John Nov 19 '13 at 20:19

The inequality is equivalent to

$$\frac{1}{ \left(1+\dfrac{1}{x_1}\right) \left(1+\dfrac{1}{x_2}\right)\left(1+\dfrac{1}{x_3}\right)\left(1+\dfrac{1}{x_4}\right)\left(1+\dfrac{1}{x_5}\right)\left(1+\dfrac{1}{x_6}\right)} \leq \frac{1}{2^{12}}$$

or

$$\sqrt{\prod \frac{x_i}{x_i+1}} \leq \frac{1}{4}$$

Now, by AM-GM we have

$$\sqrt{\prod \frac{x_i}{x_i+1}} \leq \frac{\sum \frac{x_i}{x_i+1}} {6}$$, so it suffices to prove

$$\sum \frac{x_i}{x_i+1} \leq \frac{3}{2}$$

This is equivalent to

$$6- \sum \frac{1}{x_i+1} \leq \frac{3}{2} \Leftrightarrow \sum \frac{1}{x_i+1} \geq \frac{9}{2}$$

But this is Just Cuachy Schwartz

$$6^2 \leq \left( \sum \frac{1}{x_i+1} \right) \left(\sum (x_i+1)\right)=\left( \sum \frac{1}{x_i+1} \right) \left(2+6\right)$$

• WOW! Really nice proof. – derivative Nov 19 '13 at 20:24
• how did you get from $\sum \frac{x_i}{x_i+1} \leq \frac{3}{2}$ to the next? n = 6, so you took it out from the summation? – user99638 Nov 19 '13 at 20:56
• @OrchidFibio $\frac{x_i}{x_i+1} =\frac{x_i+1-1}{x_i+1} =\frac{x_i+1}{x_i+1}-\frac{1}{x_i+1}=1-\frac{1}{x_i+1}$ – N. S. Nov 19 '13 at 20:57
• How did you get to the last step? – user99638 Nov 19 '13 at 20:57
• @OrchidFibio Added some details in my last comment... – N. S. Nov 19 '13 at 20:58

The function $$f(x):=\log\left(1+{1\over x}\right)\qquad(x>0)$$ has a positive second derivative, whence it is convex. Jensen's inequality then implies $$\sum_{k=1}^6{1\over 6}\log\left(1+{1\over x_k}\right)\geq\log\left(1+{1\over\sum_{k=1}^6{1\over6} x_k}\right)\ .\tag{1}$$ As $\sum_{k=1}^6 x_k=2$ the right side of $(1)$ evaluates to $\log 4$, so that we immediately obtain $$\prod_{k=1}^6\left(1+{1\over x_k}\right)\geq 4^6=2^{12}\ .$$

• How did you get to the last step? – user99638 Nov 19 '13 at 21:35
• $$\sum_{k=1}^6\log\left(1+{1\over x_k}\right)\geq 6 \log 4 = \log 4^6$$ then use sum of logs is the log of the products on the left, and take the antilog of both sides. – John Nov 19 '13 at 23:48

If $$\sum^{n}_{i=1}x_i=S$$ By $AM\ge GM$, we have $$1+x_i=\frac{Sx_i+\sum^{n}_{k=1}x_k}{S} \ge\frac{n+S}{S}\sqrt[n+S]{x_i^s\prod^{n}_{k=1}x_k}$$ $$\implies \frac{1+x_i}{\sqrt[n+S]{x_i^S\prod^{n}_{k=1}x_k}}\ge \frac{n+S}{S}$$ $$\implies \prod^{n}_{i=1}{\frac{1+x_i}{\sqrt[n+S]{x_i^S\prod^{n}_{k=1}x_k}}}\ge \left(\frac{n+S}{S}\right)^n$$ $$\implies \prod^{n}_{i=1}\frac{1+x_i}{x_i}\ge \left(\frac{n+S}{S}\right)^n$$ Now plug in $n=6,S=2$ and be forever happy.