Question

\(a^4+b^4\le\frac{a^6}{b^2}+\frac{b^6}{a^2}\)

Answer 1

\(\frac{a^6}{b^2}+\frac{b^6}{a^2}=\frac{a^8+b^8}{a^2b^2}\ge\frac{\left(a^4+b^4\right)^2}{2a^2b^2}=\frac{\left(a^4+b^4\right)\left(a^4+b^4\right)}{2a^2b^2}\ge\frac{2a^2b^2\left(a^4+b^4\right)}{2a^2b^2}\)

Answer 2

\(\frac{a^6}{b^2}+a^2b^2\ge2\sqrt{\frac{a^6}{b^2}.a^2b^2}=2a^4\)

\(\Rightarrow\frac{a^6}{b^2}\ge2a^4-a^2b^2\). Tương tự rồi cộng lại suy ra:

\(\frac{a^6}{b^2}+\frac{b^6}{a^2}\ge2\left(a^4+b^4\right)-2a^2b^2\)

\(\ge2\left(a^4+b^4\right)-\left(a^4+b^4\right)=a^4+b^4\)

Đẳng thức xảy ra khi a = b