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Áp dụng bđt Cauchy Schwarz dưới dạng Engel ta có :
\(\frac{\left(a+b\right)^2}{c}+\frac{\left(c+b\right)^2}{a}+\frac{\left(a+c\right)^2}{b}\ge\frac{\left(a+b+c+b+c+a\right)^2}{a+b+c}\)
\(=\frac{\left(2a+2b+2c\right)^2}{a+b+c}=\frac{4\left(a+b+c\right)^2}{a+b+c}=4\left(a+b+c\right)\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
Ta có : \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2ac+2bc=0\)
\(\Rightarrow a^2+b^2+c^2=-2ab-2bc-2ac\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=\left(-2ab-2bc-2ac\right)^2\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4\left(a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2\right)\)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+a^2c^2\right)+2abc\left(a+b+c\right)\)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+a^2c^2\right)\) (do \(a+b+c=0\))
\(\Leftrightarrow\frac{a^4+b^4+c^4}{2}=\frac{2a^2b^2+2b^2c^2+2a^2c^2}{2}\)
\(\Leftrightarrow\frac{a^4+b^4+c^4}{2}+\frac{a^4+b^4+c^4}{2}=\frac{2a^2b^2+2b^2c^2+2a^2c^2}{2}+\frac{a^4+b^4+c^4}{2}\)
\(\Leftrightarrow a^4+b^4+c^4=\frac{a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2}{2}\)
\(\Rightarrow a^4+b^4+c^4=\frac{\left(a^2+b^2+c^2\right)^2}{2}\)(đpcm)