Let \(a, b\) and \(c\) be positive real numbers such that \(a + b + c \leq 4\) and \(ab + bc + ca \geq 4\).

Prove that at least two of the inequalities

\(|a - b| \leq 2, |b - c| \leq 2, |c - a| \leq 2\)

are true.

Let \(a, b\) and \(c\) be positive real numbers such that \(a + b + c \leq 4\) and \(ab + bc + ca \geq 4\).

Prove that at least two of the inequalities

\(|a - b| \leq 2, |b - c| \leq 2, |c - a| \leq 2\)

are true.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestFor sake of contradiction let us suppose \(|a-b| > 2\), \(|b-c| > 2\),\(|c-a| > 2\). which implies \[(a-b)^{2} + (b-c)^{2} + (c-a)^{2} > 8\] ---(1)

Given \(a + b + c \le 4\) \[a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca \le 16\] \[a^{2} + b^{2} + c^{2} - ab - bc - ca \le 16 - 3ab -3bc - 3ca\] \[a^{2} + b^{2} + c^{2} - ab - bc - ca \le 4\] \[2a^{2} + 2b^{2} + 2c^{2} - 2ab - 2bc - 2ca \le 8\] \[(a-b)^{2} + (b-c)^{2} + (c-a)^{2} \le 8\].

This contradicts (1) and hence our desired proof follows. – Eddie The Head · 2 years, 11 months ago

Log in to reply

its not true – Mayank Singh · 2 years, 11 months ago

Log in to reply

– Sharky Kesa · 2 years, 11 months ago

It actually is true.Log in to reply