ta có \(\frac{1+2+3+...+2013.a}{a}\)< \(\frac{1+2+3+...+2013.b}{b}\)nên ta có

(1+2+3+...+2013.a ) : a < (1+2+3+...+2013.b) :b

vì 2013 x a chia hết cho aneen loại và 2013.b chia hết cho b nên loại . Vậy

(1+2+3+.... ) :a <(1+2+3+...):b

mà 1+2+3+... = 1+2+3+...

nên chắc chắn rằng 1+2+3+... :a vì a lớn hơn b nên 1+2+3 +...:a <1+2+3+... :

Vậy a >b

\(\frac{1+2+3+...+2013a}{a}=\frac{1+2+3+...+2013a-1}{a}+\frac{2013a}{a}=\frac{1+2+3+...+2013a-1}{a}+2013\)

\(\frac{1+2+3+...+2013b}{b}=\frac{1+2+3+...+2013b-1}{b}+\frac{2013b}{b}=\frac{1+2+3+...+2013b-1}{b}+2013\)

suy ra \(\frac{1+2+3+...+2013a-1}{a}<\frac{1+2+3+...+2013b-1}{b}\)

\(\Rightarrow\frac{2013a-1}{a}<\frac{2013b-1}{b}\Rightarrow\frac{a\left(2013-\frac{1}{a}\right)}{a}<\frac{b\left(2013-\frac{1}{b}\right)}{b}\)

\(\Rightarrow2013-\frac{1}{a}<2013-\frac{1}{b}\Rightarrow\frac{1}{a}<\frac{1}{b}\Rightarrow b>a\)

nhầm , b<a

Ta có: \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\)

=>\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}\)

Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=k\)

=>\(\begin{cases}d=ak\\ c=dk=ak\cdot k=ak^2\\ b=ck=ak^2\cdot k=ak^3\\ a=bk=ak^3\cdot k=ak^4\end{cases}\Rightarrow ak^4-a=0\)

=>\(a\left(k^4-1\right)=0\)

=>\(k^4-1=0\)

=>\(k^4=1\)

=>k=1

=>a=b=c=d

\(A=\frac{2013a-2012b}{c+d}+\frac{2013b-2012c}{a+d}+\frac{2013c-2012d}{a+b}+\frac{2013d-2012a}{b+c}\)

\(=\frac{2013a-2012a}{a+a}+\frac{2013a-2012a}{a+a}+\frac{2013a-2012a}{a+a}+\frac{2013a-2012a}{a+a}\)

\(=\frac12+\frac12+\frac12+\frac12=\frac42=2\)

\(\sum\dfrac{a}{a+\sqrt{\left(a+b\right)\left(c+a\right)}}\le\sum\dfrac{a}{a+\sqrt{ab}+\sqrt{ac}}=\sum\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\)

Ta có kết quả tổng quát hơn như sau:

Cho $a,b,c \neq 0$ thỏa mãn $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0.$

Chứng minh rằng $$S={\frac {k{a}^{2}-k-1}{{a}^{2}+2\,bc}}+{\frac {{b}^{2}k-k-1}{2\,ac+{b}^{2}}}+{\frac {{c}^{2}k-k-1}{2\,ab+{c}^{2}}}=k$$

Ta có : \(\frac{a}{a+\sqrt{2013a+bc}}=\frac{a}{a+\sqrt{a^2+ab+ac+bc}}=\frac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\)

Theo bất đẳng thức Bunhiacopxki : \(\sqrt{\left(a+b\right)\left(c+a\right)}\ge\sqrt{\left(\sqrt{ac}+\sqrt{ab}\right)^2}=\sqrt{ab}+\sqrt{ac}\)

\(\Rightarrow\frac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{a}{a+\sqrt{ab}+\sqrt{ac}}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

hay \(\frac{a}{a+\sqrt{2013a+bc}}\le\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Tương tự : \(\frac{b}{b+\sqrt{2013b+ac}}\le\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

\(\frac{c}{c+\sqrt{2013c+ab}}\le\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Cộng các bất đẳng thức trên theo vế được \(\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ac}}+\frac{c}{c+\sqrt{2013c+ab}}\le1\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\\a+b+c=2013\\a,b,c>0\end{cases}}\) \(\Leftrightarrow a=b=c=671\)

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