Ta có: \(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)+3abc\)

\(=3\left(a^2+b^2+c^2\right)-3\left(ab+bc+ac\right)+3abc\)

Xét: \(4\left(a^2+b^2+c^2\right)-\left(a^3+b^3+c^3\right)\ge9\)(1)

<=> \(\left(a^2+b^2+c^2\right)+3\left(ab+bc+ac\right)-3abc\ge9\)

<=> \(\left(a+b+c\right)^2+\left(ab+bc+ac\right)-3abc\ge9\)

<=> \(ab+bc+ac\ge3abc\)

<=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\)(2)

Để chứng (1) đúng ta cần chứng minh (2) đúng

Thật vậy ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=\frac{9}{3}=3\)

=> (2) đúng 

Vậy (1) đúng 

Dấu "=" xảy ra <=> a = b = c =1 .

Đầu tiên chứng minh:

\(a^3+b^3+c^3\ge ba^2+cb^2+ac^2\)

Ta có:

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

\(\ge3\left(a^2b+b^2c+c^2a\right)\)

\(\Rightarrow a^3+b^3+c^3\ge ba^2+cb^2+ac^2\)

Quay lại bài toán ta có:

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

\(=\frac{a^4}{a^2+a^2b-a^3}+\frac{b^4}{b^2+b^2c-b^3}+\frac{c^4}{c^2+c^2a-c^3}\)

\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)+\left(a^2b+b^2c+c^2a\right)-\left(a^3+b^3+c^3\right)}\)

\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)+\left(a^3+b^3+c^3\right)-\left(a^3+b^3+c^3\right)}\)

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

Áp dụng BĐT AM-GM ta có: 

\(\frac{a^2}{1+b-a}+a^2\left(1+b-a\right)\ge2a^2\)

\(\frac{b^2}{1+c-b}+b^2\left(1+c-b\right)\ge2b^2\)

\(\frac{c^2}{1+a-c}+c^2\left(1+a-c\right)\ge2c^2\)

Cộng theo vế 3 BĐT trên ta có:

\(VT+a^2b+b^2c+c^2a-a^3-b^3-c^3\ge1\)

Cần chứng minh \(a^3+b^3+c^3\ge a^2b+b^2c+c^2a\)

Tiếp tục xài AM-GM \(a^3+a^3+b^3\ge3\sqrt[3]{a^6b^3}=3a^2b\)

TƯơng tự rồi cộng theo vế ta có ĐPCM

Xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)

1/ Ta cần c/m: \(3x^2+3y^2+3z^2\ge x^2+y^2+z^2+2\left(xy+yz+zx\right)\)

Tức là \(2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) (đúng)

Ta có đpcm.

\(3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=12\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)

\(\Leftrightarrow\)\(4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{16}\le\frac{49}{16}\)

\(\Leftrightarrow\)\(\left[2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{4}\right]^2\le\frac{49}{16}\)

\(\Leftrightarrow\)\(\frac{-7}{4}\le2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{4}\le\frac{7}{4}\)

\(\Leftrightarrow\)\(\frac{-3}{4}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le1\)

Có : \(\frac{1}{4a+b+c}+\frac{1}{a+4b+c}+\frac{1}{a+b+4c}\le\frac{1}{36}\left(\frac{6}{a}+\frac{6}{b}+\frac{6}{c}\right)\le\frac{1}{6}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=3\)

... 

Trong ba số a,b,c không mất tính tổng quát ta có thể giả sử rằng b là số nằm giữa hai số a,c nghĩa là \(a\le b\le c\)  hoặc \(a\ge b\ge c\). Khi đó ta có \(c\left(b-a\right)\left(b-c\right)\le0\to b^2c+c^2a\le bc^2+abc\to a^2b+b^2c+c^2a\le a^2b+bc^2+abc\)

\(=b\left(a^2+c^2+ac\right)\le b\left(a+c\right)^2=\frac{1}{2}\times2\cdot2b\cdot\left(a+c\right)^2\)

\(\le\frac{1}{2}\left(\frac{2b+\left(a+c\right)+\left(a+c\right)}{3}\right)^3=\frac{1}{2}\left(\frac{2\left(a+b+c\right)}{3}\right)^3=\frac{4}{27}\)