\(\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\)
\(=\frac{a^2}{ab+ca-a^2}+\frac{b^2}{ab+bc-b^2}+\frac{c^2}{ca+bc-c^2}\)
\(\ge\frac{\left(a+b+c\right)^2}{2ab+2bc+2ca-a^2-b^2-c^2}\)
\(\ge\frac{3\left(ab+bc+ca\right)}{2ab+2bc+2ca-ab-bc-ca}=3\)
\(VT=\frac{2\left(a-b\right)^2}{\left(b+c-a\right)\left(c+a-b\right)}+\frac{2\left(b-c\right)^2}{\left(c+a-b\right)\left(a+b-c\right)}+\frac{2\left(a-c\right)^2}{\left(a+b-c\right)\left(b+c-a\right)}+3\ge3\)
Cách khác Đặt \(\hept{\begin{cases}b+c-a=x\\a+c-b=y\\a+b-c=z\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}c=\frac{x+y}{2}\\b=\frac{x+z}{2}\\a=\frac{z+y}{2}\end{cases}}\)
BĐT cần CM \(\Leftrightarrow\frac{z+y}{2x}+\frac{x+z}{2y}+\frac{x+y}{2z}\ge3\)
\(\Leftrightarrow\frac{z}{x}+\frac{y}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}\ge6\)
Đúng (do \(\frac{x}{y}+\frac{y}{x}\ge2;\frac{x}{z}+\frac{z}{x}\ge2;\frac{z}{y}+\frac{y}{z}\ge2\))
Dấu "="
Giả sử \(a=min\left\{a,b,c\right\}\)
\(VT-VP=\frac{\left(b-c\right)^2\left(4b+4c-5a\right)+a\left(b+c-2a\right)^2}{\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)}\ge0\)
Ta có đpcm.
P/s: Đây là phân tích bằng tay nên đôi khi có thể có sai sót, anh tự check ạ! Nếu sai thì nhắn em phát để em sửa lại.
Cách 5:
\(VT-VP=\frac{\left(a+b-2c\right)^2}{c\left(a+b-c\right)}+\frac{\left(a-b\right)^2\left(a+b+c\right)}{c\left(b+c-a\right)\left(c+a-b\right)}\ge0\)
Đẳng thức xảy ra khi \(a=b=c\)
Giả sử \(c=min\left\{a,b,c\right\}\)
\(VT-VP=\frac{4\left[c\left(\Sigma_{cyc}a^2-\Sigma_{cyc}ab\right)+\left(a-b\right)^2\left(a+b-2c\right)\right]}{\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}\ge0\)
\(a+b-c>0;b+c-a>0;a+c-b>0\)
\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right);a+b-c=x;b+c-a=y;c+a-b=z\)
\(\Rightarrow a=\frac{x+z}{2};b=\frac{x+y}{2};c=\frac{z+y}{2}\Rightarrow abc=\frac{y+z}{2}\cdot\frac{x+z}{2}\cdot\frac{x+y}{2}=\frac{\left(y+z\right)\left(x+z\right)\left(x+y\right)}{8}\)
\(\Leftrightarrow8xyz\ge\left(y+z\right)\left(x+z\right)\left(x+y\right)\)
\(=\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}\ge6\)\(2\left(\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\right)=\frac{2a}{b+c-a}+\frac{2b}{c+a-b}+\frac{2c}{a+b-c}\)
\(\Leftrightarrow\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\ge3\)
