\(\frac{a^2}{b}+b\ge2a;\frac{b^2}{c}+c\ge2b;\frac{c^2}{a}+a\ge2c\)(BĐT cô-si)
\(\Rightarrow\frac{a^2}{b}+b+\frac{b^2}{c}+c+\frac{c^2}{a}+a\ge2a+2b+2c\)
\(\Rightarrowđpcm\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

đặt b+c-a=x,a+c-b=y,a+b-c=z thì x,y,z>0 do a,b,c>0

=>x+y+z=a+b+c

có a=(y+z)/2 , b=(z+x)/2 ,c=(x+y)/2

A=(y+z)/2x + (z+x)/2y + (x+y)/2z =1/2[(x/y+y/x)+(y/z+z/y)+(x/z+z/x)

Áp dụng bđt cosi : x/y+y/x >= 2,y/z+z/y >= 2,z/x+x/z >= 2 

=>A >= 1/2.6=3 (đpcm)

Dấu "=" xảy ra <=> x=y=z<=>b+c-a=a+c-b=a+b-c<=>a=b=c <=> tam giác đó là tam gíac đều

Áp dụng bđt Cauchy-Schawrz dạng Engel ta có:

A = a^2/ab+ac-a^2  +  b^2/ab+bc-b^2  +  c^2/ac+bc-c^2

A \(\ge\)(a+b+c)^2/2.(ab+bc+ca)-(a^2+b^2+c^2)

A \(\ge\)a^2+b^2+c^2+2.(ab+bc+ca)/2.(ab+bc+ca)-(a^2+b^2+c^2)

A \(\ge\)2.(ab+bc+ca)-(a^2+b^2+c^2)/2.(ab+bc+ca)-(a^2+b^2+c^2)  +  2.(a^2+b^2+c^2)/2.(ab+bc+ca)-(a^2+b^2+c^2)

A \(\ge\)1  +  2.(a^2+b^2+c^2)/2.(a^2+b^2+c^2)-(a^2+b^2+c^2)

A \(\ge\) 1 + 2 = 3 (đpcm)

Dấu "=" xảy ra khi a = b = c

ta có \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}.\)

áp dụng vào bài ta có\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9>6\)

Ta có :\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(=a\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

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

\(=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)

Nhận thấy \(\frac{a}{b}+\frac{b}{a}\ge2\)

Thật vậy ta có : \(\frac{a}{b}+\frac{b}{a}\ge2\)

<=> \(\frac{a^2+b^2}{ab}\ge2\Rightarrow a^2+b^2\ge2ab\Rightarrow a^2-2ab+b^2\ge0\Rightarrow\left(a-b\right)^2\ge0\left(\text{đúng}\right)\)

Tương tự ta chứng minh được \(\hept{\begin{cases}\frac{a}{c}+\frac{c}{a}\ge2\\\frac{b}{c}+\frac{c}{b}\ge2\end{cases}}\)

Khi đó \(3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)\ge3+2+2+2\ge9>6\)(đpcm)