đề trường nào đây bạn

\(a^{2012}+b^{2012}+c^{2012}\ge3\sqrt[3]{\left(abc\right)^{2012}}=3\)

\(\Rightarrow\dfrac{1}{a^{2012}+b^{2012}+c^{2012}}\le\dfrac{1}{3}\)

\(\Rightarrow-\dfrac{1}{a^{2012}+b^{2012}+c^{2012}}\ge-\dfrac{1}{3}\)

Lại có:

\(a^{2013}+a^{2013}+...+a^{2013}\left(\text{2012 số hạng}\right)+1\ge2013\sqrt[2013]{\left(a^{2013}\right)^{2012}}=2013.a^{2012}\)

\(\Rightarrow2012.a^{2013}+1\ge2013.a^{2012}\)

Tương tự: \(2012.b^{2013}+1\ge2013.b^{2012}\) ; \(2012.c^{2013}+1\ge2013.c^{2012}\)

Cộng vế với vế:

\(\Rightarrow a^{2013}+b^{2013}+c^{2013}\ge\dfrac{2013\left(a^{2012}+b^{2012}+c^{2012}\right)-3}{2012}\)

\(\Rightarrow A\ge\dfrac{2013\left(a^{2012}+b^{2012}+c^{2012}\right)-3}{2012\left(a^{2012}+b^{2012}+c^{2012}\right)}=\dfrac{2013}{2012}-\dfrac{3}{2012}.\dfrac{1}{a^{2012}+b^{2012}+c^{2012}}\ge\dfrac{2013}{2012}-\dfrac{3}{2012}.\dfrac{1}{3}=1\)

\(A_{min}=1\) khi \(a=b=c=1\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

=> bc+ac+ab=0

ta có

\(bc+ac=-ab\)

<=> \(\left(bc+ac\right)^2=a^2b^2\)

<=> \(b^2c^2+a^2c^2+2abc^2=a^2b^2\)

<=> \(b^2c^2+a^2c^2-a^2b^2=-2abc^2\)

tương tự

\(a^2b^2+b^2c^2-c^2a^2=-2ab^2c\)

\(c^2a^2+a^2b^2-b^2c^2=-2a^2bc\)

thay vào E ta đc

\(E=\dfrac{-a^2b^2c^2}{2ab^2c}-\dfrac{a^2b^2c^2}{2abc^2}-\dfrac{a^2b^2c^2}{2a^2bc}\)

=\(-\dfrac{ac}{2}-\dfrac{ab}{2}-\dfrac{bc}{2}=\dfrac{-\left(ac+ab+bc\right)}{2}=0\) (vì ac+bc+ab=0 cmt)

P = \(\frac{a^2c}{a^2c+c^2b+b^2a+}+\frac{b^2a}{b^2a+a^2c+c^2b}+\frac{c^2b}{c^2b+b^2a+a^2c}\)

P = \(\frac{a^2c+b^2a+c^2b}{a^2c+c^2b+b^2a}=1\)

\(P=\frac{\frac{a}{b}}{\frac{a}{b}+\frac{c}{a}+\frac{b}{c}}+\frac{\frac{b}{c}}{\frac{b}{c}+\frac{a}{b}+\frac{c}{a}}+\frac{\frac{c}{a}}{\frac{c}{a}+\frac{b}{c}+\frac{a}{b}}=\frac{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}=1\)

Áp dụng BĐT cô-si, ta có 

\(a^2+\frac{1}{a^2}\ge2\sqrt{a^2.\frac{1}{a^2}}=2\)

Tương tự, ta có \(a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge6\)

dấu= xảy ra <=>\(a^2=b^2=c^2=1\)

=>\(a^{2012}=b^{2012}=c^{2012}=1\Rightarrow a^{2012}+b^{2012}+c^{2012}=3\left(ĐPCM\right)\)

^_^

tự làm

Ta có : \(a^2+b^2=c^2+d^2\)

\(\Leftrightarrow a^2-c^2=d^2-b^2\)

\(\Leftrightarrow\left(a-c\right)\left(a+c\right)=\left(d-b\right)\left(d+b\right)\)

Do \(a+b=c+d\Rightarrow a-c=d-b\)

\(\Rightarrow\left(a-c\right)\left(a+c\right)=\left(a-c\right)\left(d+b\right)\)

\(\Leftrightarrow\left(a-c\right)\left(a+c-b-d\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-c=0=d-b\\a+c=b+d\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=c\\d=b\end{matrix}\right.\\a+c=b+d\end{matrix}\right.\)

Với a = c ; d = b \(\Rightarrow a^{2012}+b^{2012}=c^{2012}+d^{2012}\left(đpcm\right)\)

Với \(a+c=b+d\)

Mà \(a+b=c+d\)

\(\Rightarrow a+c+a+b=b+d+c+d\)

\(\Rightarrow2a=2d\Rightarrow a=d\Rightarrow a^{2012}=d^{2012}\left(1\right)\)

Lại có : \(a+c=b+d\)

\(\Rightarrow b=c\Rightarrow b^{2012}=c^{2012}\left(2\right)\)

Từ ( 1 ) ; ( 2 )

\(\Rightarrow a^{2012}+b^{2012}=c^{2012}+d^{2012}\left(đpcm\right)\)