B tự c/m BĐT \(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)nhé.
Dấu " = " xảy ra \(\Leftrightarrow x=y=z\)
Áp dụng :
\(x^4+y^4+z^4\ge\frac{1}{3}.\left(x^2+y^2+z^2\right)^2\ge\frac{1}{3}.\left[\frac{1}{3}.\left(x+y+z\right)^2\right]^2=\frac{1}{27}.\left(x+y+z\right)^4=\frac{1}{27}.2^4=\frac{16}{27}\)
Dấu " = " xảy ra \(\Leftrightarrow x=y=z=\frac{2}{3}\)
KL:...
vận dụng bất đẳng thức x^2+y^2+z^2 \(\ge\) (x+y+z)^2/3
Áp dụng BĐT AM-GM ta có:
\(x^4+\frac{16}{81}+\frac{16}{81}+\frac{16}{81}\ge4.\sqrt[4]{x^4.\frac{16}{81}.\frac{16}{81}.\frac{16}{81}}=\frac{32}{27}x\)
Dấu " = " xảy ra \(\Leftrightarrow x^4=\frac{16}{81}\Leftrightarrow x=\frac{2}{3}\)
Tương tự:
\(y^4+\frac{16}{81}+\frac{16}{81}+\frac{16}{81}\ge4.\sqrt[4]{y^4.\frac{16}{81}.\frac{16}{81}.\frac{16}{81}}=\frac{32}{27}y\)
\(z^4+\frac{16}{81}+\frac{16}{81}+\frac{16}{81}\ge4.\sqrt[4]{z^4.\frac{16}{81}.\frac{16}{81}.\frac{16}{81}}=\frac{32}{27}z\)
Dấu " = " xảy ra \(\Leftrightarrow y^4=\frac{16}{81}\Leftrightarrow y=\frac{2}{3}\)
\(z^4=\frac{16}{81}\Leftrightarrow z=\frac{2}{3}\)
Cộng vế với vế của 3 BĐT trên ta có:
\(x^4+y^4+z^4+\frac{16}{81}.9\ge\frac{32}{27}\left(x+y+z\right)\)
\(\Leftrightarrow x^4+y^4+z^4\ge\frac{16}{27}\)
Dấu " = " xảy ra \(\Leftrightarrow x=y=z=\frac{2}{3}\)
Vậy Min \(x^4+y^4+z^4=\frac{16}{27}\)\(\Leftrightarrow x=y=z=\frac{2}{3}\)