nicekikah

2021-08-20

Prove that:

1.$2ab\le {a}^{2}+{b}^{2}$

2.$ab+ac+bc\le {a}^{2}+{b}^{2}+{c}^{2}$

If a, b and c are all integers.

1.

2.

If a, b and c are all integers.

comentezq

Skilled2021-08-21Added 106 answers

1. a and b are integers, therefore a-b is an integer. An integers square is always greater then or equal to 0.

$0\le {(a-b)}^{2}$

$(a-b)}^{2}={a}^{2}-2ab+{b}^{2$

$0\le {a}^{2}-2ab+{b}^{2}$

$2ab\le {a}^{2}-2ab+{b}^{2}+2ab$

$2ab\le {a}^{2}+{b}^{2}$

2. Since a, b and c are integers, any difference between them is also an integer. An integers square is always greater then or equal to 0.

$0\le {(a-b)}^{2}+{(b-c)}^{2}+{(c-a)}^{2}$

$0\le {a}^{2}-2ab+{b}^{2}+{b}^{2}-2bc+{c}^{2}+{c}^{2}-2ac+{a}^{2}$

$2ab+2bc+2ac\le 2{a}^{2}+2{b}^{2}+2{c}^{2}$

$ab+bc+ac\le {a}^{2}+{b}^{2}+{c}^{2}$

2. Since a, b and c are integers, any difference between them is also an integer. An integers square is always greater then or equal to 0.

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