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# Absolute 2

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.

Note by Sharky Kesa
2 years, 6 months ago

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For 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. · 2 years, 6 months ago

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its not true · 2 years, 6 months ago

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It actually is true. · 2 years, 6 months ago

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