+  +  ≥ 3.

Đặt b + c – a = x > 0 (1); a + c – b = y > 0  (2); a + b – c = z > 0  (3)

Cộng (1) và (2) => b + c – a + a + c – b = x + y ⇔ 2c = x + y ⇔ c = 

Tương tự a =  ; b = 

Do đó  +  +  =  +   +  = ( +  +  +  +  + )

= [( + ) + ( + ) + ( + )] ≥ (2 + 2 + 2) = 3.

Vậy  +  +  ≥ 3.

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Giải:

Ta có BĐT phụ: \(\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\le abc\)

Áp dụng BĐT Cauchy - Schwarz ta có:

\(\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}\)

\(\ge3\sqrt[3]{\dfrac{abc}{\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}}\)

\(\ge3\sqrt[3]{\dfrac{abc}{abc}}\ge3\) (Đpcm)

Thấy đún

-Đặt \(\left\{{}\begin{matrix}b+c-a=x>0\\c+a-b=y>0\\a+b-c=z>0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2c=x+y\\2a=y+z\\2b=z+x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}c=\dfrac{x+y}{2}\\a=\dfrac{y+z}{2}\\b=\dfrac{z+x}{2}\end{matrix}\right.\)

\(A=\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}=\dfrac{\dfrac{y+z}{2}}{x}+\dfrac{\dfrac{z+x}{2}}{y}+\dfrac{\dfrac{x+y}{2}}{z}=\dfrac{1}{2}\left(\dfrac{y+z}{x}+\dfrac{z+x}{y}+\dfrac{x+y}{z}\right)=\dfrac{1}{2}\left[\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)+\left(\dfrac{x}{z}+\dfrac{z}{x}\right)\right]\ge\dfrac{1}{2}.\left(2+2+2\right)=3\left(đpcm\right)\)

-Dấu "=" xảy ra khi \(a=b=c\)

Áp dụng BĐT AM-GM ta có:

\(2\sqrt{\dfrac{y+z-x}{x}}\le\dfrac{y+z-x}{x}+1=\dfrac{y+z}{x}\)

\(\Leftrightarrow\sqrt{\dfrac{x}{y+z-x}}\ge\dfrac{2x}{y+z}\)

Áp dụng vào đề bài ta có:

\(A=\sqrt{\dfrac{a}{b+c-a}}+\sqrt{\dfrac{b}{c+a-b}}+\sqrt{\dfrac{c}{a+b-c}}\ge\)

\(\ge\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}\ge2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=\dfrac{2.3}{2}=3\)(BĐT Nesbitt)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

\(VT=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ac+bc}\)

\(VT\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c\)

\(\Rightarrow\) Tam giác là tam giác đều

\(\Leftrightarrow ab\left(\dfrac{1}{b+c}-\dfrac{1}{a+c}\right)+bc\left(\dfrac{1}{a+c}-\dfrac{1}{a+b}\right)+ca\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)=0\)

\(\Leftrightarrow\dfrac{ab\left(a-b\right)}{\left(b+c\right)\left(a+c\right)}+\dfrac{bc\left(b-c\right)}{\left(a+b\right)\left(a+c\right)}+\dfrac{ca\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}=0\)

\(\Leftrightarrow\dfrac{ab\left(a^2-b^2\right)+bc\left(b^2-c^2\right)+ca\left(c^2-a^2\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\) hay tam giác cân