The sum of the three nonnegative real numbers $ x_1, x_2, x_3$ is not greater than $\frac12$. Prove that $(1 - x_1)(1 - x_2)(1 - x_3) \ge \frac12$
Problem
Source:
Tags: algebra, inequalities
sqing
16.11.2021 03:12
parmenides51 wrote: The sum of the three nonnegative real numbers $ x_1, x_2, x_3$ is not greater than $\frac12$. Prove that $(1 - x_1)(1 - x_2)(1 - x_3) \ge \frac12$ https://math.stackexchange.com/questions/2086207/an-inequality-with-a-load-of-variables-1-a-11-a-2-1-a-n-ge-1-2?noredirect=1 https://math.stackexchange.com/questions/3812834/want-some-hint-on-a-proof-of-an-inequality-by-induction?noredirect=1
brainfertilzer
17.11.2021 04:41
I feel like the following proof is wrong:
Let $p(x) = x^3 - ax^2 + bx - c$ be a polynomial with roots $x_1,x_2,x_3$. we want to prove that $p(1) = 1-a+b-c\ge \frac{1}{2}$. Note that $a = x_1 + x_2 + x_3\le\frac{1}{2}$.
It suffices to show that the minimum value of $p(1)$ is greater than or equal to $\frac{1}{2}$. This happens when $a$ and $c$ are maximized while $b$ is minimized. Note that the maximum value of $a$ is $1/2$, and the maximum value of $c$ occurs when $x_1 = x_2 = x_3 = \frac{1}{6}\implies c_{\text{max}} = \frac{1}{216}$. In this case,
\[ b = \sum_{ 3\ge i,j\ge 1, i\ne j}x_ix_j = 3\cdot\left(\frac{1}{6}\right)^2 = \frac{1}{12}.\]
Thus, $p(1)_{\text{min}} = 1 - \frac{1}{2} + \frac{1}{12} -\frac{1}{216}>\frac{1}{2}$.
Can someone please tell me if this is indeed wrong and if so, how to fix it?
greenturtle3141
17.11.2021 04:53
A good effort, but I think the issue lies in how this claim is poorly-defined: brainfertilzer wrote: This happens when $a$ and $c$ are maximized while $b$ is minimized. If $a,b,c$ were independent quantities then sure, you can maximize one with no issue of minimizing another. But they depend on each other, and as such obtaining the maximum of one of them may prevent you from obtaining the extremum of another.
brainfertilzer
17.11.2021 05:11
Oh, I get it now. Thanks @above.