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Ta có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=4a^2+4b^2+4c^2-4ab-4bc-4ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=4a^2+4b^2+4c^2-4ab-4bc-4ac\)
\(\Leftrightarrow0=2a^2+2b^2+2c^2-2ab-2bc-2ac\)
\(\Leftrightarrow0=a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2\)
\(\Leftrightarrow0=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
Mà \(\left\{\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)
\(\Rightarrow a=b=c\) ( đpcm )
\(2=\left(a^2+ab+\dfrac{b^2}{4}\right)+\left(a^2-2+\dfrac{1}{a^2}\right)-ab\)
\(2=\left(a+\dfrac{b}{2}\right)^2+\left(a-\dfrac{1}{a}\right)^2-ab\ge-ab\)
\(\Rightarrow ab\ge-2\)
Dấu "=" xảy ra khi \(\left(a;b\right)=\left(1;-2\right);\left(-1;2\right)\)
\(a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2\)
\(\Leftrightarrow a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2=0\)
\(\Leftrightarrow\left(a^4-2a^2b^2+b^4\right)+\left(b^4-2b^2c^2+c^4\right)+\left(c^4-2c^2a^2+a^4\right)-a^4-b^4-c^4=0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(c^2-b^2\right)^2+\left(c^2-a^2\right)^2-a^4-b^4-c^4=0\)
\(\Leftrightarrow\left(a-b\right)^2c^2+a^2\left(b+c\right)^2+b^2\left(c+a\right)^2-a^4-b^4-c^4=0\)
\(\Leftrightarrow c^2\left[\left(a-b\right)^2-\left(a+b\right)^2\right]+a^2\left[\left(b+c\right)^2-a^2\right]+b^2\left[\left(c+a\right)^2-b^2\right]=0\)
\(\Leftrightarrow c^2\left[\left(a-b\right)^2-\left(a+b\right)^2\right]+a^2\left[\left(b+c\right)^2-\left(c-b\right)^2\right]+b^2\left[\left(c+a\right)^2-\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow-4abc^2+4a^2bc+4ab^2c=0\)
\(\Leftrightarrow4abc\left(a+b-c\right)=0\)
\(\Leftrightarrow0=0\)(luôn đúng)
=>đpcm
chứng minh j vậy bạn
a^4 + b^4 = (a^2 + b^2)^2 - 2a^2 . b^2