\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Leftrightarrow ab+bc+ca=0\)

\(\left(a+b+c\right)^2=1\Leftrightarrow a^2+b^2+c^2+2.\left(ab+bc+ca\right)=1\)

\(\Leftrightarrow a^2+b^2+c^2+2.0=1\)

\(\Leftrightarrow a^2+b^2+c^2=1\)

2. Áp dụng bất đẳng thức Cô - si cho 3 số dương \(\frac{a}{b},\frac{b}{c},\frac{c}{a}\)ta có

\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\sqrt[3]{\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}\)\(=3\)

Dấu "=" xảy ra <=> a = b = c

Ta biến đổi 1 tí nhé

\(\frac{4}{a}+\frac{5}{b}+\frac{3}{c}\ge4\left(\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{c+a}\right)\)

\(\Leftrightarrow\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{a+c}\le\frac{1}{4}\left(\frac{4}{a}+\frac{5}{b}+\frac{3}{c}\right)\)

Tới đây dễ dàng áp dụng BĐT \(\frac{4}{x+y}\le\frac{1}{x}+\frac{1}{y}\)

\(\Leftrightarrow\frac{3}{a+b}\le\frac{3}{4}.\frac{1}{a}+\frac{3}{4}.\frac{1}{b}\left(1\right)\)

\(\Leftrightarrow\frac{2}{b+c}\le\frac{1}{2}.\frac{1}{b}+\frac{1}{2}.\frac{1}{c}\left(2\right)\)

\(\Leftrightarrow\frac{1}{a+c}\le\frac{1}{4}.\frac{1}{a}+\frac{1}{4}.\frac{1}{c}\left(3\right)\)

Cộng vế với vế của (1), (2), (3) suy ra 

\(\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{a+c}\le\frac{3}{4}\cdot\frac{1}{a}+\frac{3}{4}\cdot\frac{1}{b}+\frac{1}{2}\cdot\frac{1}{b}+\frac{1}{2}\cdot\frac{1}{c}+\frac{1}{4}\cdot\frac{1}{a}+\frac{1}{4}\cdot\frac{1}{c}\)

\(\Leftrightarrow\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{a+c}\le\frac{1}{a}+\frac{5}{4}\cdot\frac{1}{b}+\frac{3}{4}\cdot\frac{1}{b}\)

\(\Leftrightarrow\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{a+c}\le\frac{1}{4}\left(\frac{4}{a}+\frac{5}{b}+\frac{3}{c}\right)\)

\(\Leftrightarrow Dpcm\)

thay a^3+b^3=(a+b)^3 -3ab(a+b) .Ta có :

a^3+b^3+c^3-3abc=0

<=>(a+b)^3 -3ab(a+b) +c^3 - 3abc=0

<=>[(a+b)^3 +c^3] -3ab.(a+b+c)=0

<=>(a+b+c). [(a+b)^2 -c.(a+b)+c^2] -3ab(a+b+c)=0

<=>(a+b+c).(a^2+2ab+b^2-ca-cb+c^2-3ab)...

<=>(a+b+c).(a^2+b^2+c^2-ab-bc-ca)=0

luôn đúng do a+b+c=0

Đề: Cho  \(a+b+c=1\) và  \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)  .  Chứng minh:  \(a^2+b^2+c^2=1\)

                                                                 -----------------------------------------

Từ   \(a+b+c=1\)

\(\Rightarrow\)  \(\left(a+b+c\right)^2=1\)

\(\Leftrightarrow\)  \(a^2+b^2+c^2+2\left(ab+bc+ca\right)=1\)  \(\left(1\right)\)

Mặt khác, ta lại có:   \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)  \(\Leftrightarrow\)  \(\frac{ab+bc+ca}{abc}=0\)  \(\Leftrightarrow\)  \(ab+bc+ca=0\)  \(\left(2\right)\)

Từ  \(\left(1\right)\)  và  \(\left(2\right)\), suy ra  \(a^2+b^2+c^2=1\)   \(\left(đpcm\right)\)