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Let $a,b,c,d$ be real numbers such that $a^2+b^2+c^2+d^2=1$. Prove that $$(1-a)(1-b)(1-c)(1-d)\geq abcd.$$

I thought about substituting $a=\sqrt{w},b=\sqrt{x}$, etc. (assuming first that $a,b,c,d$ are positive), and then looking at the convexity of the function $f(r)=(1-\sqrt{r})/\sqrt{r}$ and applying some Jensen-type inequality. But such an inequality applies to the sum of functions, not the product.

Alexi
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1 Answers1

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Observe that as long as $a,b,c,d \ge 0$: $$\begin{align}(1-a)(1-b) - cd &\ge 1 - a - b + ab - \left(\frac{c^2+d^2}{2}\right) \\ &= 1 - a - b + ab - \left(\frac{1-a^2-b^2}{2}\right) \\ &= \frac{1}{2}(1-a-b)^2 \ge 0\end{align}$$

Similarly, $$(1-c)(1-d) -ab \ge \frac{1}{2}(1-c-d)^2\ge 0$$

Hence, $$(1-a)(1-b)(1-c)(1-d) \ge abcd$$

Przemysław Scherwentke
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r9m
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    what if $cd < 0$, and $ab < 0$ ? example: $4 > -2, 5 > -11$, but $20 < (-2)(-11) = 22$. – DeepSea Dec 18 '14 at 04:03
  • @ah-huh-moment. oh noes ! I was hurrying ! Didn't notice the OP is missing a positivity constraint ! (I edit accordingly) – r9m Dec 18 '14 at 04:06
  • If some of $a,b,c,d$ is negative, then $abcd$ never goes up, while $(1-a)(1-b)(1-c)(1-d)$ always goes up or stays the same from the corresponding positive values of $a,b,c,d$. So your proof still works :) – Alexi Dec 18 '14 at 04:14