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\(\Sigma_{sym}a^4b^4\ge\frac{\left(\Sigma_{sym}a^2b^2\right)^2}{3}\ge\frac{\left(\Sigma_{sym}ab\right)^4}{27}\ge\frac{a^2b^2c^2\left(a+b+c\right)^2}{3}=3a^4b^4c^4\)
\(\Sigma\frac{a^5}{bc^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{abc\left(a+b+c\right)}\ge\frac{\left(a^2+b^2+c^2\right)^4}{abc\left(a+b+c\right)^3}\ge\frac{\left(a+b+c\right)^6\left(a^2+b^2+c^2\right)}{27abc\left(a+b+c\right)^3}\)
\(\ge\frac{\left(3\sqrt[3]{abc}\right)^3\left(a^2+b^2+c^2\right)}{27abc}=a^2+b^2+c^2\)
#)Giải :
a) Để C/m a và b là hai số đối nhau => a + b = 0
Ta có : \(2\left(a^2+b^2\right)=\left(a-b\right)^2\)
\(\Leftrightarrow2a^2+2b^2=a^2-2ab+b^2\)
\(\Leftrightarrow2a^2+2b^2-a^2-2ab+b^2=0\)
\(\Leftrightarrow a^2+2ab+b^2=0\)
\(\Leftrightarrow\left(a+b\right)^2=0a\Leftrightarrow a+b=0\)
\(\Rightarrowđpcm\)
Ta có: \(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ca\right)\right]^2\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2=4\left[a^2b^2+b^2c^2+2abc\left(a+b+c+\right)\right]\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4a^2b^2+4b^2c^2+4c^2a^2\) (a + b + c = 0)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\) (1)
Mà \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2\) (2)
Từ (1) và (2) \(\Rightarrow\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+a^4+b^4+c^4=2\left(a^4+b^4+c^4\right)\)
=> đpcm
Ta co: \(a^4+b^4+c^4\)
\(=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(=\left(a^2+b^2+c^2\right)^2-2\left[\left(ab+bc+ca\right)^2-2\left(ab^2c+abc^2+a^2bc\right)\right]\)
\(=\left(a^2+b^2+c^2\right)^2-2\left[\left(ab+bc+ca\right)^2-2abc\left(a+b+c\right)\right]\)
\(=\left(a^2+b^2+c^2\right)^2-2\left(ab+bc+ca^2\right)\)
\(\Rightarrow2\left(a^4+b^4+c^4\right)=2\left(a^2+b^2+c^2\right)^2-4\left(ab+bc+ca\right)^2\)
\(\Rightarrow2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2-\left(2ab+2bc+2ca\right)^2\) \(\left(1\right)\)
Lại có: \(a+b+c=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)
\(\Rightarrow2ab+2bc+2ca=-\left(a^2+b^2+c^2\right)\)
\(\Rightarrow\left(2ab+2bc+2ca\right)^2=\left(a^2+b^2+c^2\right)^2\) \(\left(2\right)\)
Từ (1) và (2) suy ra
\(2\left(a^4+b^4+c^4\right)=2\left(a^2+b^2+c^2\right)^2-\left(a^2+b^2+c^2\right)^2\)
\(=\left(a^2+b^2+c^2\right)^2\)