\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
Ta c/m BĐT phụ: \(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2\)( b tự c/m nhé. Chuyển vế, c/m VP>=0 là xong )
\(\Rightarrow\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{\left(a+b+c\right)^2}{2.\frac{1}{3}\left(a+b+c\right)^2}=\frac{1}{\frac{2}{3}}=\frac{3}{2}\)
\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
đpcm
\(ab+bc+ca\le\frac{1}{3}.\left(a+b+c\right)^2\)
\(\Leftrightarrow3.\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\)
\(\Leftrightarrow3.\left(ab+bc+ca\right)\le a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)( BĐT luôn đúng)
\(\Rightarrow ab+bc+ca\le\frac{1}{3}.\left(a+b+c\right)^2\)
đpcm
Không mất tính tổng quát chuẩn hóa a+b+c=3
Ta có: \(\frac{a\left(b+c\right)}{4}\)+\(\frac{a}{b+c}\)>=a
tương tự ta đc:P>=a+b+c-\(\frac{ab+bc+ca}{2}\)>=a+b+c-\(\frac{\left(a+b+c\right)^2}{6}\)=3-9/6=3/2 Dấu bằng xảy ra <=> a=b=c=1 hay a=b=c
Không mất tính tổng quát giả sử \(a\le b\le c\Rightarrow a+b\le b+c\le c+a\)
\(\Rightarrow\hept{\begin{cases}a\le b\le c\\\frac{1}{b+c}\le\frac{1}{c+a}\le\frac{1}{a+b}\end{cases}}\)Sử dụng BĐT hoán vị ta có:
\(\hept{\begin{cases}\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}\\\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{c}{c+b}+\frac{a}{c+a}+\frac{b}{a+b}\end{cases}}\)
\(\Rightarrow\frac{2a}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b}\ge\frac{b+c}{b+c}+\frac{c+a}{c+a}+\frac{a+b}{a+b}\ge\frac{3}{2}\left(đpcm\right)\)
Dấu "=" <=> a=b=c
