\(sigma\frac{a}{1+b^2}=sigma\left(a-\frac{ab^2}{1+b^2}\right)\ge sigma\left(a\right)-sigma\frac{ab}{2}\ge3-\frac{\frac{\left(a+b+c\right)^2}{3}}{2}=\frac{3}{2}>\frac{2018}{2003}\)

Quy đồng full lên, hồi sáng e làm bên  H O C 2 4 rồi, giờ chả muốn nhai lại.

anh cx hỏi câu đấy mà :))

Ta co:

\(a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}\le\frac{ab+ca}{2}+\frac{bc+ab}{2}+\frac{ca+bc}{2}=ab+bc+ca\)

Suy ra BDT can phai chung minh la:

\(a^2+b^2+c^2\ge ab+bc+ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(dung)

Dau '=' xay khi \(a=b=c\)

A=\(\left(1+\frac{\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a-1}}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)

a rút gọn A

b tìm A sao cho A>1

C tính a biết a=2020-2 căn2019

=\(\frac{1}{\sqrt{5}}+\frac{5\left(2\sqrt{5}+5\sqrt{5}\right)}{\left(2\sqrt{5}+5\sqrt{5}\right)\left(2\sqrt{5}-5\sqrt{5}\right)}+\frac{2\left(2+\sqrt{5}\right)}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}\)

=\(\frac{1}{\sqrt{5}}-7\sqrt{5}-2\sqrt{5}-4\)=\(\frac{1-45-4\sqrt{5}}{\sqrt{5}}=\frac{-44\sqrt{5}-20}{5}\)