It is known about numbers $$a_1, a_2, ... a_n$$ that $$a_1 + a_2 +...+a_n \le 1/2.$$ Prove that $$ (1-a_1)(1-a_2)...(1-a_n) \ge 1/2$$ I have tried using $a^2 \geq 0$, it led to nothing. How can I make my inequality look like in the possible duplicate?
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Arithmetic Mean - Geometric Mean inequality (AM-GM). – mathreadler Jan 06 '17 at 14:46
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3@mathreadler How is the AM-GM mean work here if we're not given the numbers are non-negative? – DonAntonio Jan 06 '17 at 14:57
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3Isn't that wrong for $a_1 = 1, a_2 = -1/2$ ? Or are all numbers assumed to be non-negative? – Martin R Jan 06 '17 at 15:00
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Yes Martin is right. It is wrong if we can not assume non-negativity. – mathreadler Jan 06 '17 at 15:01
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The inequality seems wrong; consider $f(x)=(x-a_1)(x-a_2)....(x-a_n)$. It is ONLY given that the sum of roots is less than or equal to $\frac{1}{2}$.. using this information no conclusion can be made about the value/range of $f(1)=(1-a_1)(1-a_2)...(1-a_n)$ – Serenity Jan 06 '17 at 15:03
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For non-negative numbers you can easily prove by induction that $(1-a_1)(1-a_2)...(1-a_n) \ge 1 - (a_1 + a_2 +...+a_n)$. But that must have been answered before ... – Martin R Jan 06 '17 at 15:05
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Arnaldo: all numbers can't be lower than the arithmetic average – mathreadler Jan 06 '17 at 15:06
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@mathreadler: Sure! My bad! – Arnaldo Jan 06 '17 at 15:13
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And can you advise me how to make it look like AM-GM? – iwannasolveit Jan 06 '17 at 15:14
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Or how to make it look like that? (1−a1)(1−a2)...(1−an)≥1−(a1+a2+...+an) By the way, this is the Bernoulli's inequality – iwannasolveit Jan 06 '17 at 15:15
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I am trying to check now what my trail of thoughts were. – mathreadler Jan 06 '17 at 15:15
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See also How to show $x_1,x_2, \dots ,x_n \geq 0 $ and $ x_1 + x_2 + \dots + x_n \leq \frac{1}{2} \implies (1-x_1)(1-x_2) \cdots (1-x_n) \geq \frac{1}{2}$. Found using Approach0. – Martin Sleziak Jan 06 '17 at 15:57
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By induction for $n=1$ it is obivious. You can consider $b_n = a_n+a_{n+1}$ where one has $$a_1, a_2, ... a_n$$ that $$a_1 + a_2 +...+\underbrace{a_n +a_{n+1}}_{b_n}\le 1/2 .$$But $$(1-a_n)(1-a_{n+1}) = (1-a_n-a_{n+1} + a_n a_{n+1} )\geq (1-a_n-a_{n+1})= (1-b_n).$$
Then $$(1-a_1)(1-a_2)...(1-a_n)(1-a_{n+1})\ge (1-a_1)(1-a_2)...(1-b_n) \ge 1/2$$
Guy Fsone
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Hint: you can consider the inequality for arithmetic and geometric means:
$$\frac{1}{n}\sum_{k=1}^n b_i \geq \sqrt[n]{\prod_{k=1}^n b_i}$$
for any set of non-negative real numbers $\{b_1,\cdots,b_n\}$
Partial attempt:
$$b_i = 1-a_i$$ $$\frac{1}{n}\sum_{k=1}^n (1-a_i) \geq \sqrt[n]{\prod_{k=1}^n (1-a_i)}$$ $$1-\frac{1}{n}\sum a_i \geq \sqrt[n]{\prod_{k=1}^n (1-a_i)}$$
$$lhs > 1-\frac{1}{2n}$$
if product $\leq 1/2$ then rhs not sure to be larger than $\sqrt[n]{\frac{1}{2}}$
Now what to remains is to compare which can be done by calculus comparing a linear function with a power function, treating n as continous variable.
The proof that blue does not overtake red can be left as an exercise.
mathreadler
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2Just to be clear: before giving your hint, you did check that it worked, right? (It may just be me, but I don't really see how this helps) – Clement C. Jan 06 '17 at 14:59
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@ClementC Okay, I have given some sketch. Please point out anything unrealistic that you find. – mathreadler Jan 06 '17 at 16:01
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I am about to head out, but quickly: since $\sum_i b_i = \sum_(1-a_i) = n - \sum_i a_i \geq n/2$, isn't the inequalty written "lhs" in the wrong direction? – Clement C. Jan 06 '17 at 16:04
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the sum is subtracted, so something smaller than $\frac{1}{2n}$ is being subtracted. – mathreadler Jan 06 '17 at 16:09
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No, the way it is written $\frac{1}{n}\sum_i b_i \geq 1-\frac{1}{2n}$, since you defined $b_i=1-a_i$ and $\frac{1}{n}\sum_i a_i \leq \frac{1}{2n}$. – Clement C. Jan 06 '17 at 16:10
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Yes wait maybe you are right, the b:s should be a:s and vice versa. So I edited so the $b_i$ are in the inequality and being substituted for $1-a_i$ instead. – mathreadler Jan 06 '17 at 16:16
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(From my phone) i don't see how the last bit of your suggested proof goes. Given the inequalities we have and want, why does the rhs have to be greater than $1/2^{1/n} $? – Clement C. Jan 06 '17 at 17:00
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Monotonicity of exponential function together with the $1/2$ in the value for of the product. Elementary calculus. – mathreadler Jan 06 '17 at 17:17
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I don't think this addresses what I didn't get. You have (1) $lhs \geq \sqrt[n]{\prod_{i=1}^n (1-a_i)}$ and (2) $lhs > 1-\frac{1}{2n}$. Why would $1-\frac{1}{2n} < \sqrt[n]{\frac{1}{2}}$ bring any contradiction? – Clement C. Jan 06 '17 at 18:28