- Bổ đề: Cho \(a,b,c>0\). Ta có:
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\left(1\right)\) (BĐT Nesbitt 3 số).
Chứng minh: Ta có:
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\)
- Áp dụng bất đẳng thức Caushy-Schwarz dạng Engel, ta có:
\(\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\ge\dfrac{\left(a+b+c\right)^2}{ab+ac+bc+bc+ca+cb}=\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
hay \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\left(2\right)\)
- Mặt khác: \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}\left(3\right)\)
- Từ (2) và (3) ta có:
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2.\dfrac{\left(a+b+c\right)^2}{3}}=\dfrac{3}{2}\left(đpcm\right)\)
- Vậy bất đẳng thức \(\left(1\right)\) đã được chứng minh. Dấu "=" xảy ra khi \(a=b=c\).
- Quay trở lại bài toán. Ta có:
\(A=\dfrac{\sqrt{a^2+1}}{b+c}+\dfrac{\sqrt{b^2+1}}{c+a}+\dfrac{\sqrt{c^2+1}}{a+b}\)
\(=\dfrac{\sqrt{\left(a^2+1\right)\left(1+3\right)}}{\sqrt{\left(1+3\right)}.\left(b+c\right)}+\dfrac{\sqrt{\left(b^2+1\right)\left(1+3\right)}}{\sqrt{\left(1+3\right)}.\left(c+a\right)}+\dfrac{\sqrt{\left(c^2+1\right)\left(1+3\right)}}{\sqrt{\left(1+3\right)}.\left(a+b\right)}\)
- Áp dụng bất đẳng thức Bunhiacopxki, ta có:
\(A\ge\dfrac{a.1+1.\sqrt{3}}{2\left(b+c\right)}+\dfrac{b.1+1.\sqrt{3}}{2\left(c+a\right)}+\dfrac{c.1+1.\sqrt{3}}{2\left(a+b\right)}\)
\(=\dfrac{1}{2}\left(\dfrac{a+\sqrt{3}}{b+c}+\dfrac{b+\sqrt{3}}{c+a}+\dfrac{c+\sqrt{3}}{a+b}\text{}\text{}\text{}\right)\)
\(=\dfrac{1}{2}\left[\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+\sqrt{3}.\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\right]\left(4\right)\)
- Áp dụng bất đẳng thức Caushy-Schwarz dạng Engel, ta có:
\(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\ge\dfrac{\left(1+1+1\right)^2}{b+c+c+a+a+b}=\dfrac{9}{2\left(a+b+c\right)}\ge\dfrac{9}{2\sqrt{3}}=\dfrac{3\sqrt{3}}{2}\left(5\right)\)- Từ (1), (4) và (5), ta có:
\(A\ge\dfrac{1}{2}\left(\dfrac{3}{2}+\sqrt{3}.\dfrac{3\sqrt{3}}{2}\right)=3\left(đpcm\right)\)
- Dấu "=" xảy ra khi \(a=b=c=\dfrac{\sqrt{3}}{3}\)
