Question

Cho \(a^2+b^2=1\). Tìm GTNN của \(a^6+b^6\)

Giúp mik vs TT !!! Cho 2 tick lun

Answer 1

\(a^6+b^6=\left(a^2\right)^3+\left(b^2\right)^3=\)

\(=\left(a^2+b^2\right)\left(a^4-a^2b^2+b^4\right)=1.\left(\left(a^2+b^2\right)^2-3a^2b^2\right)\)

\(=1-3a^2b^2\le1\)

vậy GTNN là 1

Answer 2

\(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2-2ab\ge0\Rightarrow2ab\le1\)(*)

\(a^6+b^6=\left(a^2\right)^{^3}+\left(b^2\right)^{^3}=\left(a^2+b^2\right)^{^3}-3a^2b^2\left(a^2+b^2\right)=1-3\left(ab\right)^2\)(**)

(*)&(**)\(a^6+b^6\ge1-3\left(\frac{1}{2}\right)^2=1-\frac{3}{4}=\frac{1}{4}\) đẳng thức khi   \(a=b=+-\frac{\sqrt{2}}{2}\)