Inequality

Bài toán : Cho các số dương $a,b,c$. Chứng minh rằng :

$\sqrt[8]{\frac{a^8+b^8}{2}}+\sqrt[8]{\frac{b^8+c^8}{2}}+\sqrt[8]{\frac{c^8+a^8}{2}}$$\le \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$

Lời giải :

Theo BĐT AM-GM :

$8\sqrt[8]{\frac{a^8+b^8}{2}}$$=8\sqrt[8]{\frac{(a^4+b^4{)}^2-2a^4{b}^4}{2}}$$=8\sqrt[8]{\frac{(a^4+b^4-a^2{b}^2\sqrt{2})(a^4+b^4+a^2{b}^2\sqrt{2})}{2}}$

$$=8\sqrt[8]{\frac{\left(a^2+b^2-ab\sqrt{2+\sqrt{2}}\right).\left(a^2+b^2+ab\sqrt{2+\sqrt{2}}\right).\left(a^2+b^2-ab\sqrt{2-\sqrt{2}}\right).\left(a^2+b^2+ab\sqrt{2-\sqrt{2}}\right).b^4}{\left(2-\sqrt{2+\sqrt{2}}\right).\left(2+\sqrt{2+\sqrt{2}}\right).\left(2-\sqrt{2-\sqrt{2}}\right).\left(2+\sqrt{2-\sqrt{2}}\right).b^4}}$$

$$\le \frac{a^2+b^2-ab\sqrt{2+\sqrt{2}}}{b\left(2-\sqrt{2+\sqrt{2}}\right)}+\frac{a^2+b^2+ab\sqrt{2+\sqrt{2}}}{b\left(2+\sqrt{2+\sqrt{2}}\right)}+\frac{a^2+b^2-ab\sqrt{2-\sqrt{2}}}{b\left(2-\sqrt{2-\sqrt{2}}\right)}+\frac{a^2+b^2+ab\sqrt{2-\sqrt{2}}}{b\left(2+\sqrt{2-\sqrt{2}}\right)}+b+b+b+b$$

$$=\frac{8a^2}{b}+12b-12a$$

Suy ra :

$$\sqrt[8]{\frac{a^8+b^8}{2}}\le \frac{a^2}{b}+\frac{3}{2}(b-a)$$

Xây dựng các BĐT tương tự rồi cộng vế theo vế, ta được điều phải chứng minh.