Prove that: 1. 2ab<=a^2+b^2 2. ab+ac+bc<=a^2+b^2+c^2 If a, b and c are all

nicekikah 2021-08-20 Answered
Prove that:
1. \(\displaystyle{2}{a}{b}\le{a}^{{2}}+{b}^{{2}}\)
2. \(\displaystyle{a}{b}+{a}{c}+{b}{c}\le{a}^{{2}}+{b}^{{2}}+{c}^{{2}}\)
If a, b and c are all integers.

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Expert Answer

comentezq
Answered 2021-08-21 Author has 10206 answers
1. a and b are integers, therefore a-b is an integer. An integers square is always greater then or equal to 0.
\(\displaystyle{0}\le{\left({a}-{b}\right)}^{{2}}\)
\(\displaystyle{\left({a}-{b}\right)}^{{2}}={a}^{{2}}-{2}{a}{b}+{b}^{{2}}\)
\(\displaystyle{0}\le{a}^{{2}}-{2}{a}{b}+{b}^{{2}}\)
\(\displaystyle{2}{a}{b}\le{a}^{{2}}-{2}{a}{b}+{b}^{{2}}+{2}{a}{b}\)
\(\displaystyle{2}{a}{b}\le{a}^{{2}}+{b}^{{2}}\)
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.
\(\displaystyle{0}\le{\left({a}-{b}\right)}^{{2}}+{\left({b}-{c}\right)}^{{2}}+{\left({c}-{a}\right)}^{{2}}\)
\(\displaystyle{0}\le{a}^{{2}}-{2}{a}{b}+{b}^{{2}}+{b}^{{2}}-{2}{b}{c}+{c}^{{2}}+{c}^{{2}}-{2}{a}{c}+{a}^{{2}}\)
\(\displaystyle{2}{a}{b}+{2}{b}{c}+{2}{a}{c}\le{2}{a}^{{2}}+{2}{b}^{{2}}+{2}{c}^{{2}}\)
\(\displaystyle{a}{b}+{b}{c}+{a}{c}\le{a}^{{2}}+{b}^{{2}}+{c}^{{2}}\)
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