chuyenndu

giả sử $\sqrt{2}=\frac{a}{b}$ với (a,b)=1

$a^2=2b^2\Rightarrow a^2\vdots 2\Rightarrow a\vdots 2\Rightarrow a=2c\Rightarrow b^2=2c^2\Rightarrow ...\Rightarrow b\vdots 2$

nên $(a,b)\vdots 2$ (VL)

$a^2+2bc+1=a^2+2bc+\frac{\sum a^2}{2}=a^2+bc+\frac{a^2+(b+c)^2}{2}\ge a^2+bc+ab+ac=(a+b)(a+c)$

$\Rightarrow \sum\frac{a^2}{a^2+2bc+1}\le \sum\frac{a^2}{(a+b)(a+c)}$

$\sum\frac{a^2}{(a+b)(a+c)}\le 1\Leftrightarrow 2abc\ge 0$

y chang vmo 1997

https://artofproblem...h224593p1247384

Walker's inequality

https://artofproblem...d_its_extension

https://artofproblem...1165552p5567071