lỗi hả

ta có \(\frac{a}{1+b-a}+a\left(1+b-a\right)\ge2a\)hay \(\frac{a}{1+b-a}\ge a\left(1+a-b\right)=a\left(2a+c\right)\)

tương tự ta sẽ có :

\(\frac{a}{1+b-a}+\frac{b}{1+c-b}+\frac{c}{1+a-c}\ge2a^2+2b^2+2c^2+ab+ac+bc\)

\(\ge\frac{3}{2}\left(a^2+b^2+c^2\right)+\frac{1}{2}\left(a^2+b^2+c^2+2ab+2bc+2ac\right)\ge\frac{1}{2}\left(a+b+c\right)^2+\frac{1}{2}\left(a+b+c\right)^2\)

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

vậy ta  có điều phải chứng minh

dấu bằng xảy ra khi \(a=b=c=\frac{1}{3}\)

vì bạn muốn làm bằng BDT Bunhia nên mình làm cách đó nhé : 

ta có : \(\left[a\left(1+b-a\right)+b\left(1+c-b\right)+c\left(1+a-c\right)\right]\left(\frac{a}{1+b-a}+\frac{b}{1+c-b}+\frac{c}{1+a-c}\right)\)

\(\ge\left(a+b+c\right)^2=1\) ( áp dụng Bunhia ) 

nên ta có : \(VT\ge\frac{1}{a\left(1+b-a\right)+b\left(1+c-b\right)+c\left(1+a-c\right)}=\frac{1}{a\left(2b+c\right)+b\left(2c+a\right)+c\left(2a+c\right)}\)

\(\ge\frac{1}{3\left(ab+bc+ca\right)}\) mà \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=\frac{1}{3}\)

nên ta có : \(VT\ge\frac{1}{3\times\frac{1}{3}}=1=VP\) vậy ta có đpcm

Ta có \(\sqrt{9+4\sqrt{2}}=\sqrt{8+2.2\sqrt{2}+1}=\sqrt{\left(2\sqrt{2}\right)^2+2.2\sqrt{2}.1+1}=\sqrt{\left(2\sqrt{2}+1\right)^2}=2\sqrt{2}+1\)

Vậy \(\sqrt{2+\sqrt{9+4\sqrt{2}}}=\sqrt{2+\left(2\sqrt{2}+1\right)}=\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}\)\(=\sqrt{2}+1\)

Từ đó \(\sqrt{5-2\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{5-2\left(\sqrt{2}+1\right)}=\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}\)\(=\sqrt{2}-1\)

Vậy \(T=\frac{\sqrt{2}-1}{\sqrt{2}-1}=1\), vậy ta có đpcm