Thôi đang rảnh, giúp bạn bài này luôn vậy!!
Giải:
Ta có:
\(VT=\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+\left(\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}+\dfrac{a^2}{a+b}\right)=A+B\)
\(A+3=\dfrac{1}{2}\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left[\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right]\)
\(\ge\dfrac{1}{2}3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}3\sqrt[3]{\dfrac{1}{a+b}\dfrac{1}{b+c}\dfrac{1}{c+a}}=\dfrac{9}{2}\)
\(\Rightarrow A\ge\dfrac{3}{2}\)
\(1^2=\left(a+b+c\right)^2\le\left(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\right)\left(a+b+b+c+c+a\right)\)
\(\Leftrightarrow1\le B.2\Leftrightarrow B\ge\dfrac{1}{2}\)
Từ đó ta có: \(VT\ge\dfrac{3}{2}+\dfrac{1}{2}=2=VP\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
\(\dfrac{a+b^2}{b+c}+\dfrac{b+c^2}{c+a}+\dfrac{c+a^2}{a+b}\ge2\)
\(\Leftrightarrow\dfrac{a\left(a+b+c\right)+b^2}{b+c}+\dfrac{b\left(a+b+c\right)+c^2}{c+a}+\dfrac{c\left(a+b+c\right)+a^2}{a+b}\ge2\)
\(\Leftrightarrow\dfrac{a^2+ab+ac+b^2}{b+c}+\dfrac{ab+b^2+bc+c^2}{c+a}+\dfrac{ca+bc+c^2+a^2}{a+b}\ge2\)
\(\Leftrightarrow\dfrac{a^2+b^2+a\left(b+c\right)}{b+c}+\dfrac{b^2+c^2+b\left(c+a\right)}{c+a}+\dfrac{c^2+a^2+c\left(a+b\right)}{a+b}\ge2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{b+c}+\dfrac{b^2+c^2}{c+a}+\dfrac{c^2+a^2}{a+b}+1\ge2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{b+c}+\dfrac{b^2+c^2}{c+a}+\dfrac{c^2+a^2}{a+b}\ge1\)
\(\Leftrightarrow\dfrac{\sqrt{\left(a^2+b^2\right)^2}}{b+c}+\dfrac{\sqrt{\left(b^2+c^2\right)^2}}{c+a}+\dfrac{\sqrt{\left(c^2+a^2\right)^2}}{a+b}\ge1\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Leftrightarrow\dfrac{\sqrt{\left(a^2+b^2\right)^2}}{b+c}+\dfrac{\sqrt{\left(b^2+c^2\right)^2}}{c+a}+\dfrac{\sqrt{\left(c^2+a^2\right)^2}}{a+b}\ge\dfrac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2\left(a+b+c\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{\left(a^2+b^2\right)^2}}{b+c}+\dfrac{\sqrt{\left(b^2+c^2\right)^2}}{c+a}+\dfrac{\sqrt{\left(c^2+a^2\right)^2}}{a+b}\ge\dfrac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2}\)
Áp dụng bất đẳng thức Mincopski
\(\Rightarrow\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\ge\sqrt{2\left(a+b+c\right)^2}=\sqrt{2}\)
\(\Rightarrow\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2\ge2\)
\(\Rightarrow\dfrac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2}\ge1\)
\(\Rightarrow\dfrac{\sqrt{\left(a^2+b^2\right)^2}}{b+c}+\dfrac{\sqrt{\left(b^2+c^2\right)^2}}{c+a}+\dfrac{\sqrt{\left(c^2+a^2\right)^2}}{a+b}\ge1\)
\(\Leftrightarrow\dfrac{a+b^2}{b+c}+\dfrac{b+c^2}{c+a}+\dfrac{c+a^2}{a+b}\ge2\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)
