Áp dụng t/c dãy tỉ số bằng nhau, ta có:

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

Suy ra

a + b = 2c

b + c = 2a

a + c = 2b

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

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

    =\(\frac{2c}{b}.\frac{2a}{c}.\frac{2b}{a}\)

    =\(\frac{8abc}{abc}\)

    = 8

Ta có: \(a+b+c=1\Rightarrow\hept{\begin{cases}a=1-b-c\\b=1-a-c\\c=1-a-b\end{cases}}\)

\(\Rightarrow\left(ab+c\right)\left(bc+a\right)\left(ac+b\right)\)\(=\left(ab+1-a-b\right)\left(bc+1-b-c\right)\left(ac+1-a-c\right)\)

\(=\left[\left(ab-a\right)-\left(b-1\right)\right]\left[\left(bc-b\right)-\left(c-1\right)\right]\left[\left(ac-c\right)-\left(a-1\right)\right]\)

\(=\left[a\left(b-1\right)-\left(b-1\right)\right]\left[b\left(c-1\right)-\left(c-1\right)\right]\left[c\left(a-1\right)-\left(a-1\right)\right]\)

\(=\left(a-1\right)\left(b-1\right)\left(c-1\right)\left(b-1\right)\left(a-1\right)\left(c-1\right)\)

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

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

\(\text{Vì }a+b+c=2014\)

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

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

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

\(\Rightarrow\left(a+b\right).\left(\frac{1}{ab}+\frac{1}{ca+bc+c^2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a+b=0\\\frac{1}{ab}+\frac{1}{ac+bc+c^2}=0\end{cases}\Rightarrow\orbr{\begin{cases}a=-b\\ab+ac+bc+c^2=0\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}a=-b\\\left(a+c\right).\left(b+c\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}a=-b\\a=-c\end{cases}\text{hoặc }b=-c}}\)

Thay vào M, ta có:

Th1: \(a=-b\Rightarrow M=\frac{1}{-b^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=\frac{1}{c^{2013}}\)

Th2: \(a=-c\Rightarrow M=\frac{1}{-c^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=\frac{1}{b^{2013}}\)

Th3:\(b=-c\Rightarrow M=\frac{1}{a^{2013}}+\frac{1}{-c^{2013}}+\frac{1}{c^{2013}}=\frac{1}{a^{2013}}\)

Vậy ...

Ta có: \(a^2+b^2+c^2=1\)

\(\Rightarrow\left\{{}\begin{matrix}\left|a\right|\le1\\\left|b\right|\le1\\\left|c\right|\le1\end{matrix}\right.\)

Ta lại có:

\(a^3+b^3+c^3=a^2+b^2+c^2\)

\(\Leftrightarrow a^2\left(1-a\right)+b^2\left(1-b\right)+c^2\left(1-c\right)=0\)

Vì \(\left\{{}\begin{matrix}1-a\ge0\\1-b\ge0\\1-c\ge0\end{matrix}\right.\)

\(\Rightarrow a^2\left(1-a\right)+b^2\left(1-b\right)+c^2\left(1-c\right)\ge0\)

Dấu = xảy ra khi: \(\left(a,b,c\right)=\left(1,0,0;0,1,0;0,0,1\right)\)

\(\Rightarrow S=1\)

Câu 1 : áp dụng BĐT SVAC ta có \(A\ge\frac{(a+b+c)^2}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c}}=\frac{1.\sqrt{2a+2b+2c}}{\sqrt{2.}(\sqrt{b+c}+\sqrt{a+b}+\sqrt{a+c})}\)

mặt khác lại có \(\frac{\sqrt{2a+2b+2c}}{\sqrt{2}.(\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c})}\ge\frac{\sqrt{(\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c})^2}}{\sqrt{2}.\sqrt{3}.(\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c})}=\frac{1}{\sqrt{6}}\)theo bđt svac

\(\Rightarrow A\ge\frac{1}{\sqrt{6}}\)dấu bằng xảy ra tại a=b=c=\(\frac{1}{3}\)

Cần chứng minh: \(\frac{19b^3-a^3}{ab+5b^2}\le4b-a\)

Thật vậy: \(\frac{19b^3-a^3}{ab+5b^2}\le4b-a\Leftrightarrow\left(4b-a\right)\left(ab+5b^2\right)-19b^3+a^3\ge0\)

\(\Leftrightarrow4ab^2+20b^3-a^2b-5ab^2-19b^3+a^3\ge0\)

\(\Leftrightarrow\left(a^3+b^3\right)-ab\left(a+b\right)\ge0\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\)(đúng)

"=" khi a=b

Tương tự: \(\frac{19c^3-b^3}{bc+5c^2}\le4c-b;\frac{19a^3-c^3}{ac+5a^2}\le4a-c\)

Cộng theo vế: 

\(\frac{19b^3-a^3}{ab+5b^2}+\frac{19c^3-b^3}{bc+5c^2}+\frac{19a^3-c^3}{ac+5a^2}\le4b-a+4c-b+4a-c=3\left(a+b+c\right)=3\)

Dấu "=" xảy ra khi a=b=c=1/3

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

Áp dụng BĐT cosi

\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\ge\frac{3}{4}a\)

Tương tự 

=> \(A\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{2}\left(a+b+c\right)=\frac{1}{4}\left(a+b+c\right)\)

Lại có \(\left(a+b+c\right)\ge\frac{9}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{9}{1}=9\)

=> \(A\ge\frac{9}{4}\)

MinA=9/4 khi a=b=c=3

\(A=\sqrt{2x^2-4x+3}+3\)

Ta có: \(2x^2-4x+3\)

\(=2\left(x^2-2x+\frac{3}{2}\right)\)

\(=2\left(x^2-2.x.1+1^2+\frac{1}{2}\right)\)

\(=2[\left(x-1\right)^2+\frac{1}{2}]\)

\(=2\left(x-1\right)^2+1\ge1\)

\(\Rightarrow\sqrt{2\left(x-1\right)^2+1}\ge\sqrt{1}\)

\(\Rightarrow\sqrt{2\left(x-1\right)^2+1}+3\ge3+\sqrt{1}=4\)

\(\Rightarrow MinA=4\Leftrightarrow x=1\)