Prove ${a}^{4}+{b}^{4}\ge {a}^{3}b+a{b}^{3}$ for any $(a,b)\in \mathbb{R}$

Shea Stuart
2022-06-29
Answered

Prove ${a}^{4}+{b}^{4}\ge {a}^{3}b+a{b}^{3}$ for any $(a,b)\in \mathbb{R}$

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Kaylie Mcdonald

Answered 2022-06-30
Author has **19** answers

Alternatively using AM-GM inequality, we have

$\frac{a+b}{2}\ge \sqrt{ab}$

or

$(a+b{)}^{2}\ge 4ab\ge ab$

so that

$(a+b{)}^{2}-ab={a}^{2}+{b}^{2}+ab\ge 0$

$\frac{a+b}{2}\ge \sqrt{ab}$

or

$(a+b{)}^{2}\ge 4ab\ge ab$

so that

$(a+b{)}^{2}-ab={a}^{2}+{b}^{2}+ab\ge 0$

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