ta có:

\(A^2=\left(\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\right)^2\le\left(a+b+c\right)\left(\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\right)\) (BĐT Bu-nhi-a)

=>\(A^2\le\sqrt{3}\left(\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\right)\)      (*)

mặt khác ta có: \(a^2+1\ge2a\) (BĐT cauchy ) =>\(\frac{a}{a^2+1}\le\frac{1}{2}\)

tương tự ta có: \(\frac{b}{b^2+1}\le\frac{1}{2}\)    ;    \(\frac{c}{c^2+1}\le\frac{1}{2}\)

=> \(\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\le\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}\)     (**)  

từ (*),(**) => \(A^2\le\sqrt{3}.\frac{3}{2}=\frac{3\sqrt{3}}{2}\)

=>\(A\le\sqrt{\frac{3\sqrt{3}}{2}}\)

=> GTLN của A là \(\sqrt{\frac{3\sqrt{3}}{2}}\)   <=> a=b=c<\(\frac{\sqrt{3}}{3}\)

Ta có:

\(\frac{a}{\sqrt{a^2+1}}=\frac{a}{\sqrt{a^2+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}}\)

\(\le\frac{\sqrt[8]{27}a}{\sqrt{4\sqrt[4]{a^2}}}=\frac{\sqrt[8]{27a^6}}{2}\)

\(=\frac{\sqrt{3}}{2}.\sqrt[8]{a^6.\frac{1}{3}}\)

\(\le\frac{\sqrt{3}}{2}.\frac{6a+\frac{2}{\sqrt{3}}}{8}\left(1\right)\)

Tương tự ta cũng có:

\(\hept{\begin{cases}\frac{b}{\sqrt{b^2+1}}\le\frac{\sqrt{3}}{2}.\frac{6b+\frac{2}{\sqrt{3}}}{8}\left(2\right)\\\frac{c}{\sqrt{c^2+1}}\le\frac{\sqrt{3}}{2}.\frac{6c+\frac{2}{\sqrt{3}}}{8}\left(3\right)\end{cases}}\)

Từ (1), (2), (3) 

\(\Rightarrow A\le\frac{\sqrt{3}}{2}.\left(\frac{6}{8\sqrt{3}}+\frac{6}{8}\left(a+b+c\right)\right)\)

\(\le\frac{\sqrt{3}}{2}.\left(\frac{3}{4\sqrt{3}}+\frac{3\sqrt{3}}{4}\right)=\frac{3}{2}\)

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

Ta có \(\sqrt{a^2-ab+b^2}=\sqrt{\frac{1}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}\ge\sqrt{\frac{1}{4}\left(a+b\right)^2}=\frac{1}{2}\left(a+b\right)\)

=> \(\frac{1}{\sqrt{a^2-ab+b^2}}\le\frac{1}{\frac{1}{2}\left(a+b\right)}=\frac{2}{a+b}\le\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

Chứng minh tương tự, rồi cộng lại, ta có 

A\(\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)

dấu = xảy ra <=> a=b=c=1

^_^