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Ta có a+b+c=0 => \(a+b=-c\Rightarrow\left(a+b\right)^3=-c^3\Rightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=3ab\)

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

\(a^6+b^6+c^6=\left(a^3\right)^2+\left(b^3\right)^2+\left(c^3\right)^2=\left(a^3+b^3+c^3\right)^2-2\left(a^3b^3+b^3c^3+c^3a^3\right)\)

\(ab+bc+ca=0\Rightarrow a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)

Do đó: \(a^6+b^6+c^6=\left(3abc\right)^2-2\cdot3a^2b^2c^2=3a^2b^2c^2\)

Vậy \(\frac{a^6+b^6+c^6}{a^3+b^3+c^3}=\frac{3a^2b^2c^2}{3abc}=abc\left(đpcm\right)\)

Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)

=> x+y+z=0

Có \(x^3+y^3+z^3-3xyz\)

=\(\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)

=\(\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2-3xy\right]\)

=0( do x+y+z=0)

=> \(x^3+y^3+z^3=3xyz\)

<=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

\(\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=0\)

\(\Leftrightarrow x+y+z=0\)

Ta có 

\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Rightarrow x^3+y^3+z^3=3xyz\)

=> ĐPCM

Mạnh Hùng hỏi được rồi á

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

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

   \(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=0\)

   \(\Leftrightarrow\frac{a+b+c}{abc}=0\)

Mà \(a,b,c\)là số nguyên khác 0 \(\Rightarrow\)\(abc\ne0\)\(\Rightarrow\)\(a+b+c=0\)\(\Rightarrow a+b=-c\)

Ta lại có: \(a^3+b^3+c^3=\left(a+b\right)^3+c^3-3ab\left(a+b\right)\)

                                          \(=\left(a+b+c\right)^3-3.\left(a+b\right).c.\left(a+b+c\right)-3ab\left(a+b\right)\)

                                          \(=0-0-3ab\left(-c\right)\)

                                          \(=3abc⋮3\)

Vậy \(a^3+b^3+c^3=3abc⋮3\)\(\Leftrightarrow\)\(a+b+c=0\)

Thay a+b+c=2017 vào \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2017}\)  ta có:

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

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)

\(\Rightarrow\frac{a+b}{ab}+\frac{a+b+c-c}{c\left(a+b+c\right)}=0\)\(\Rightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)

\(\Rightarrow\left(a+b\right)\left(\frac{1}{ab}+\frac{1}{c\left(a+b+c\right)}\right)=0\)\(\Rightarrow\left(a+b\right)\left(\frac{c\left(a+b+c\right)+ab}{abc\left(a+b+c\right)}\right)=0\)

\(\Rightarrow\left(a+b\right)\left(\frac{c\left(b+c\right)+ca+ab}{abc\left(a+b+c\right)}\right)=0\)

\(\Rightarrow\left(a+b\right)\left[c\left(b+c\right)+ca+ab\right]=0\)

\(\Rightarrow\left(a+b\right)\left[c\left(b+c\right)+a\left(b+c\right)\right]=0\)

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

\(\Rightarrow\)\(a+b=0\) hoặc \(b+c=0\) hoặc \(c+a=0\)

\(\Rightarrow\)\(c=2017\)hoặc \(a=2017\) hoặc \(b=2017\left(đpcm\right)\)

1. Đề thiếu

2. BĐT cần chứng minh tương đương:

\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)

Ta có:

\(a^4+b^4+c^4\ge\dfrac{1}{3}\left(a^2+b^2+c^2\right)^2\ge\dfrac{1}{3}\left(ab+bc+ca\right)^2\ge\dfrac{1}{3}.3abc\left(a+b+c\right)\) (đpcm)

3.

Ta có:

\(\left(a^6+b^6+1\right)\left(1+1+1\right)\ge\left(a^3+b^3+1\right)^2\)

\(\Rightarrow VT\ge\dfrac{1}{\sqrt{3}}\left(a^3+b^3+1+b^3+c^3+1+c^3+a^3+1\right)\)

\(VT\ge\sqrt{3}+\dfrac{2}{\sqrt{3}}\left(a^3+b^3+c^3\right)\)

Lại có:

\(a^3+b^3+1\ge3ab\) ; \(b^3+c^3+1\ge3bc\) ; \(c^3+a^3+1\ge3ca\)

\(\Rightarrow2\left(a^3+b^3+c^3\right)+3\ge3\left(ab+bc+ca\right)=9\)

\(\Rightarrow a^3+b^3+c^3\ge3\)

\(\Rightarrow VT\ge\sqrt{3}+\dfrac{6}{\sqrt{3}}=3\sqrt{3}\)

4.

Ta có:

\(a^3+1+1\ge3a\) ; \(b^3+1+1\ge3b\) ; \(c^3+1+1\ge3c\)

\(\Rightarrow a^3+b^3+c^3+6\ge3\left(a+b+c\right)=9\)

\(\Rightarrow a^3+b^3+c^3\ge3\)

5.

Ta có:

\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\) ; \(\dfrac{a}{b}+\dfrac{c}{a}\ge2\sqrt{\dfrac{c}{b}}\) ; \(\dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\)

\(\Rightarrow\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}+\sqrt{\dfrac{a}{c}}\le\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=1\)

Câu 1:

\(VT=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(VT=1-\dfrac{1}{n}< 1\) (đpcm)

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

\(\Leftrightarrow\left(ab+ac+bc\right)\left(a+b+c\right)-abc=0\)

\(\Leftrightarrow\left(b+c\right)\left(ab+ac+bc\right)+a\left(ab+ac+bc\right)-abc=0\)

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

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

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

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

=> a=-b hoặc b=-c hoặc c = -a

Không mất tình tổng quát, giả sử a=-b -> a^n = -b^n ( n lẻ):

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