Inequalities

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lgx57

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lgx57

#1 h Mar 19, 2025, 7:43 AM

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Let $0 < a,b,c < 1$ . Prove that

$a(1-b)+b(1-c)+c(1-a)<1$

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invisibleman

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invisibleman

#2 h Mar 19, 2025, 8:36 AM

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Seeing the expressions, we can rightly think of Jensen's inequality. Let $f(x)=x-x^2$ and since $f"(x)=-2<0$ the function is concave, so based on Jensen's inequality
$\frac{f(a)+f(b)+f(c)}{3}\le f\left( \frac{a+b+c}{3} \right)$ that is, it remains to check that
$\frac{a+b+c}{3}-{{\left( \frac{a+b+c}{3} \right)}^{2}}<1$ which is obvious, since $x^2-x+1>0$ for all real $x$ .

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sqing

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sqing

#3 h Mar 19, 2025, 8:44 AM

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lgx57 wrote:

Let $0 < a,b,c < 1$ . Prove that

$a(1-b)+b(1-c)+c(1-a)<1$

All-Russian Olympiad 1990
https://artofproblemsolving.com/community/c6h494969p3275740
https://artofproblemsolving.com/community/c6h494492p11835829

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lgx57

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lgx57

#4 h Apr 16, 2025, 7:41 AM

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invisibleman wrote:

Seeing the expressions, we can rightly think of Jensen's inequality. Let $f(x)=x-x^2$ and since $f"(x)=-2<0$ the function is concave, so based on Jensen's inequality
$\frac{f(a)+f(b)+f(c)}{3}\le f\left( \frac{a+b+c}{3} \right)$ that is, it remains to check that
$\frac{a+b+c}{3}-{{\left( \frac{a+b+c}{3} \right)}^{2}}<1$ which is obvious, since $x^2-x+1>0$ for all real $x$ .

How do you change $a(1-b)+b(1-c)+c(1-b)$ to $a(1-a)+b(1-b)+c(1-c)$ ?

If you use rearrangement inequality, WLOG $a \le b \le c$ ,then:
$a(1-b)+b(1-c)+c(1-a)>a(1-a)+b(1-b)+c(1-c)$

This post has been edited 1 time. Last edited by lgx57, Apr 16, 2025, 7:43 AM
Reason: Wrong

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pooh123

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pooh123

#5 h Apr 16, 2025, 1:45 PM

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How about this different approach:

We have:
$a(1 - b) + b(1 - c) + c(1 - a) = (1 - b)(1 - c) + a(b + c - 1) - 1.$
Since $0 < b, c < 1$ , we have $1 - b > 0$ and $1 - c > 0$ , so
$(1 - b)(1 - c) < 1.$
Also, $a(b + c - 1) \leq |a||b + c - 1| \leq 1 \cdot 1 = 1$ , so
$(1 - b)(1 - c) + a(b + c - 1) - 1 < 1 + 1 - 1 = 1.$

This post has been edited 1 time. Last edited by pooh123, Apr 16, 2025, 1:45 PM
Reason: typo

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jjmmxx

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jjmmxx

#6 h Apr 18, 2025, 11:08 AM

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0 < (1-a)(1-b)(1-c) = 1 - (a+b+c) + (ab+bc+ca) - abc < 1 - (a+b+c-ab-bc-ca).

Therefore, a(1-b)+b(1-c)+c(1-a) = a+b+c-ab-bc-ca < 1.

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