The return of a legend inequality

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The return of a legend inequality

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Source: Legacy of Hungary

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#1 h Mar 18, 2025, 4:09 PM

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Given that $0<a,b,c,d<1$ .Prove that: $(1-a)(1-b)(1-c)(1-d) > 1-a-b-c-d$

This post has been edited 1 time. Last edited by giangtruong13, Mar 19, 2025, 11:28 AM

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#2 h Mar 18, 2025, 4:48 PM

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Using inequality $(1-x)(1-y) = 1-x-y+xy > 1-x-y$ we get:
$(1-a)(1-b)(1-c)(1-d) > (1-a-b)(1-c-d) > 1-a-b-c-d$

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polishedhardwoodtable

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polishedhardwoodtable

#3 h Mar 18, 2025, 5:01 PM

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Replace each variable with one minus itself to get $abcd-a-b-c-d+3>0$ . Then observe that $abcd-d=d(abc-1)>abc-1$ (since $abc-1<0$ ), and similarly $abc-c>ab-1$ and $ab-b>a-1$ so we have $abcd-a-b-c-d+3>abc-a-b-c+2>ab-a-b+1>0$

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arqady

#4 h Mar 18, 2025, 6:03 PM

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giangtruong13 wrote:

Given that $0<a,b,c,d<1$ .Prove that: $(1-a)(1-b)(1-c)(1-d) > 1-a-b-c-d$

This inequality is a linear inequality of any variables...

This post has been edited 1 time. Last edited by arqady, Mar 18, 2025, 6:05 PM

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#5 h Mar 18, 2025, 6:15 PM • 2 Y

Y by HotSinglesInYourArea, Calamarul

Denote $x_1=1-a, x_2=1-b, x_3=1-c, x_4=1-d$ .

The inequality rewrites as $x_1x_2x_3x_4>x_1+x_2+x_3+x_4-3$ , where $x_i\in(0,1)$ .

We will prove the following by induction: $\prod_{1\leq i\leq n}x_i>\sum_{1\leq i\leq n}x_i+1-n$ , where $n\geq 2$ and $x_i\in(0,1)$ .

Notice that for $n=2$ the inequality is $x_1x_2>x_1+x_2-1\iff (x_1-1)(x_2-1)>0$ , which is true.

Suppose the inequality is true for any $n$ numbers $x_1, ..., x_n\in(0,1)$ and prove that it is true for $n+1$ :

Denote $p=\prod_{1\leq i\leq n}x_i$ and $s=\sum_{1\leq i\leq n}x_i$ . Notice that the following inequalities are true: $0<s<n$ , $0<p<1$ and $p>s+1-n$ .

We want to prove that $px_{n+1}>x_{n+1}+s+1-n\iff x_{n+1}<\frac{n-s}{1-p}$ , for any $x_{n+1}\in(0,1)$ .

This is equivallent to proving that $1\leq \frac{n-s}{1-p}\iff 1-p\leq n-s\iff p\geq s+1-n$ , which we know it's true, so we're done.

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#7 h Mar 18, 2025, 8:51 PM • 1 Y

Y by Double07

It is true for $n$ variables which is called Weierstrass's Inequality

$\prod_{i=1}^{n}{(1-a_i)}\geq 1-\sum_{i=1}^{n}{a_i}$ holds for all $0\leq a_i\leq 1$ .

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sqing

#8 h Mar 19, 2025, 2:31 AM

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giangtruong13 wrote:

Given that $0<a,b,c,d<1$ .Prove that: $(1-a)(1-b)(1-c)(1-d) > 1-a-b-c-d$

https://artofproblemsolving.com/community/c6h1142864p5376277

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sqing

#9 h Mar 19, 2025, 2:42 AM

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Let $a,b,c,d \in[0,1]$ . Show that
$4\geq (1-a)(1-b)(1-c)(1-d) +a+b+c+d\geq 1$ Let $a_1,a_2,...,a_n\in[0,1]$ $(n\geq2)$ . Show that
$n \geq (1-a_1)(1-a_2)\cdots(1-a_n)+a_1+a_2+\cdots+a_n\geq 1$

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