 # 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.

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself. comentezq
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.
$$\displaystyle{0}\le{\left({a}-{b}\right)}^{{2}}+{\left({b}-{c}\right)}^{{2}}+{\left({c}-{a}\right)}^{{2}}$$
$$\displaystyle{0}\le{a}^{{2}}-{2}{a}{b}+{b}^{{2}}+{b}^{{2}}-{2}{b}{c}+{c}^{{2}}+{c}^{{2}}-{2}{a}{c}+{a}^{{2}}$$
$$\displaystyle{2}{a}{b}+{2}{b}{c}+{2}{a}{c}\le{2}{a}^{{2}}+{2}{b}^{{2}}+{2}{c}^{{2}}$$
$$\displaystyle{a}{b}+{b}{c}+{a}{c}\le{a}^{{2}}+{b}^{{2}}+{c}^{{2}}$$