Cho x;y>0 thỏa x+y=1. Tìm Min A=16(x4+y4)Ta có: \(A=16\left(x^4+y^4\right)\ge\frac{16.\left(x^2+y^2\right)^2}{2}=8\left(x^2+y^2\right)^2\) \(\ge8.\left[\frac{\left(x+y\right)^2}{2}\right]^2=\frac{8.\left(x+y\right)^4}{2}=2\left(x+y\right)^4=1\)Dấu =...
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Cho x;y>0 thỏa x+y=1. Tìm Min A=16(x4+y4)
Ta có: \(A=16\left(x^4+y^4\right)\ge\frac{16.\left(x^2+y^2\right)^2}{2}=8\left(x^2+y^2\right)^2\)
\(\ge8.\left[\frac{\left(x+y\right)^2}{2}\right]^2=\frac{8.\left(x+y\right)^4}{2}=2\left(x+y\right)^4=1\)
Dấu = khi \(x=y=\frac{1}{2}\)
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