\(\frac{m}{a}+\frac{n}{b}+\frac{p}{c}=0\Rightarrow mbc+nac+pab=0\)

\(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a^2=\left(b+c\right)^2=b^2+2bc+c^2\\b^2=a^2+2ac+c^2\\c^2=a^2+2ab+b^2\end{matrix}\right.\)

\(\Rightarrow A=m\left(b^2+c^2\right)+n\left(a^2+c^2\right)+p\left(a^2+b^2\right)+2\left(anp+bmp+cmn\right)\)

\(=a^2\left(n+p\right)+b^2\left(m+p\right)+c^2\left(m+n\right)\)

\(=-ma^2-nb^2-cp^2=-A\)

\(\Rightarrow A=-A\Rightarrow2A=0\Rightarrow A=0\)

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)

\(\Leftrightarrow a^2b+ab^2+c^2a+ca^2+b^2c+bc^2+2abc=0\)

\(\Leftrightarrow\left(a^2+2ab+b^2\right)c+ab\left(a+b\right)+c^2\left(a+b\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca+c^2\right)=0\)

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

=> Hoặc a+b=0 hoặc b+c=0 hoặc c+a=0

=> Hoặc a=-b hoặc b=-c hoặc c=-a

Ko mất tổng quát, g/s a=-b

a) Ta có: vì a=-b thay vào ta được:

\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{1}{b^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{c^3}\)

\(\frac{1}{a^3+b^3+c^3}=\frac{1}{-b^3+b^3+c^3}=\frac{1}{c^3}\)

=> đpcm

b) Ta có: \(a+b+c=1\Leftrightarrow-b+b+c=1\Rightarrow c=1\)

=> \(P=-\frac{1}{b^{2021}}+\frac{1}{b^{2021}}+\frac{1}{c^{2021}}=\frac{1}{1^{2021}}=1\)

Ta có:

\(\frac{x}{a}+\frac{y}{b}=\frac{x+y}{c}\)

\(\Leftrightarrow\frac{x}{a}+\frac{y}{b}=\frac{x+y}{-a-b}\)

\(\Leftrightarrow x\left(b^2+2ab\right)+y\left(a^2+2ab\right)=0\left(1\right)\)\

Ta cần chứng minh:

\(xa^2+yb^2=\left(x+y\right)c^2\)

\(\Leftrightarrow xa^2+yb^2=\left(x+y\right)\left(a+b\right)^2\)

\(\Leftrightarrow x\left(b^2+2ab\right)+y\left(a^2+2ab\right)=0\left(2\right)\)

Từ (1) và (2) ta có ĐPCM

\(\frac{d}{b^2}\) hay \(\frac{b^2}{d}\)hả bạn?

Ta có: \(\frac{a^4}{c}+\frac{b^4}{d}\ge\frac{\left(a^2+b^2\right)^2}{c+d}=\frac{1}{c+d}\)

Dấu = xảy ra khi \(\frac{a^2}{c}=\frac{b^2}{d}\)

Do đó: \(VT=\frac{a^2}{c}+\frac{b}{d^2}=\frac{d^2}{b}+\frac{b}{d^2}\ge2\sqrt{\frac{d^2}{b}.\frac{b}{d^2}}=2\)