Áp dụng bất đẳng thức Cauchy-Schwarz ta có:

\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{bc+ab}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\ge\frac{3\left(ab+bc+ac\right)}{2\left(ab+bc+ac\right)}=\frac{3}{2}\)

Áp dụng BĐT AM - GM : \(ab< \frac{a^2+b^2}{2}\Rightarrow a^2+ab+b^2\le\frac{3}{2}\left(a^2+b^2\right)\)

\(\Rightarrow\frac{a^3}{a^2+ab+b^2}\ge\frac{2}{3}.\frac{a^3}{a^2+b^2}=\frac{2}{3}\left(a-\frac{ab^2}{a^2+b^2}\right)\)

Mà cũng theo BĐT AM - GM : \(\frac{ab^2}{a^2+b^2}\le\frac{ab^2}{2ab}=\frac{b}{2}\)

\(\Rightarrow\frac{a^3}{a^2+ab+b^2}\ge\frac{2}{3}\left(a-\frac{ab^2}{a^2+b^2}\right)\ge\frac{2}{3}\left(a-\frac{b}{2}\right)\)

Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế :

\(\Rightarrow\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+a^2}\ge\frac{2}{3}\left(a-\frac{b}{2}\right)+\frac{2}{3}\left(b-\frac{c}{2}\right)\) \(+\frac{2}{3}\left(c-\frac{a}{2}\right)\)

Ta có đpcm 

Dấu " = " xảy ra khi \(a=b=c\)

Chúc bạn học tốt !!!

Tham khảo tại đây:

Câu hỏi của Đỗ Tiến Dũng - Toán lớp 9 - Học toán với OnlineMath

Lời giải:

Áp dụng BĐT AM-GM: \(ab\leq \frac{a^2+b^2}{2}\Rightarrow a^2+ab+b^2\leq \frac{3}{2}(a^2+b^2)\)

\(\Rightarrow \frac{a^3}{a^2+ab+b^2}\geq \frac{2}{3}.\frac{a^3}{a^2+b^2}=\frac{2}{3}\left(a-\frac{ab^2}{a^2+b^2}\right)\)

Mà cũng theo BĐT AM-GM: \(\frac{ab^2}{a^2+b^2}\leq \frac{ab^2}{2ab}=\frac{b}{2}\)

\(\Rightarrow \frac{a^3}{a^2+ab+b^2}\geq \frac{2}{3}\left(a-\frac{ab^2}{a^2+b^2}\right)\geq \frac{2}{3}(a-\frac{b}{2})\)

Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:

\(\Rightarrow \frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+a^2}\geq \frac{2}{3}(a-\frac{b}{2})+\frac{2}{3}(b-\frac{c}{2})+\frac{2}{3}(c-\frac{a}{2})=\frac{a+b+c}{3}\)

Ta có đpcm.

Dấu "=" xảy ra khi $a=b=c$

Ta có:\(\frac{a^3}{a^2+ab+b^2}=\frac{a\left(a^2+ab+b^2\right)-ab\left(a+b\right)}{a^2+ab+b^2}=a-\frac{ab\left(a+b\right)}{a^2+ab+b^2}\)

Lại có:\(a^2+ab+b^2\ge3ab\)

\(\Rightarrow a-\frac{ab\left(a+b\right)}{a^2+ab+b^2}\ge a-\frac{ab\left(a+b\right)}{3ab}=a-\frac{a+b}{3}\)

\(\Rightarrow\sum\frac{a^3}{a^2+ab+b^2}\ge\frac{a+b+c}{3}\)

"="<=>a=b=c

Lời giải:
BĐT đã cho tương đương với:

\(\frac{a}{b}-\frac{b}{a}+\frac{b}{c}-\frac{c}{b}+\frac{c}{a}-\frac{a}{c}\geq 0\)

\(\Leftrightarrow \frac{a^2-b^2}{ab}+\frac{b^2-c^2}{bc}+\frac{c^2-a^2}{ca}\geq 0\)

\(\Leftrightarrow \frac{a^2-b^2}{ab}-\frac{(a^2-b^2)+(c^2-a^2)}{bc}+\frac{c^2-a^2}{ca}\geq 0\)

\(\Leftrightarrow (a^2-b^2)\left(\frac{1}{ab}-\frac{1}{bc}\right)+(c^2-a^2)\left(\frac{1}{ca}-\frac{1}{bc}\right)\geq 0\)

\(\Leftrightarrow (a^2-b^2)(c-a)+(c^2-a^2)(b-a)\geq 0\)

\(\Leftrightarrow (a-b)(a+b)(c-a)-(c-a)(c+a)(a-b)\geq 0\)

\(\Leftrightarrow (a-b)(b-c)(c-a)\geq 0\) (luôn đúng với mọi $0< a\leq b\leq c$)

Ta có đpcm.

Dấu "=" xảy ra khi $a=b$ hoặc $b=c$ hoặc $c=a$

giả sử a\(\le\)b \(\le\)c.

khi đó \(\frac{a}{b+c}\le\frac{b}{c+a}\le\frac{c}{a+b}\)

áp dụng BĐT Trê bư sép ta có:

\(\left(a^2+b^2+c^2\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\le3\left(\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\right)=3VT\)

lại có a² + b² + c² \(\ge\) \(\frac{\left(a+b+c\right)^2}{3}\) nên:

3VT \(\ge\frac{\left(a+b+c\right)^2}{3}\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)

hay VT \(\ge\left(\frac{a+b+c}{3}\right)^2\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\). đpcm

\(\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)

\(\ge\left(a+b+c\right)\left(\frac{9}{b+c+c+a+a+b}\right)=\frac{\left(a+b+c\right)9}{2\left(a+b+c\right)}=\frac{9}{2}\)

\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{9}{2}-3=\frac{3}{2}\)

\(VT=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

\(=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1-3\)

\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}-3\)

\(=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

\(=\frac{1}{2}[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)]\)\(\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)

C/m BĐT phụ  \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\text{(*)   }\)  với x, y, z  dương

   Áp dụng BĐT Cô-si ta có:

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{\frac{1}{xyz}}=9\)

ÁP dụng  BĐT (*) ta có:

\(VT=\frac{1}{2}\left[\left(x+y\right)+\left(y+z\right)+\left(z+x\right)\right]\)\(\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)-3\)

\(VT\ge\frac{1}{2}.9-3=\frac{3}{2}\left(đpcm\right)\)

\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^2}{a\left(b+c\right)}+\frac{b^2}{b\left(c+a\right)}+\frac{c^2}{c\left(a+b\right)}\)

>= \(\frac{\left(a+b+c\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}=\frac{a^2+b^2+c^2+2ab+2ac+2bc}{2ab+2ac+2bc}\)(bđt cauchy schawatz dạng engel)(1)

mà \(a^2+b^2+c^2>=ab+ac+bc\)

\(\Rightarrow\frac{a^2+b^2+c^2+2ab+2ac+2bc}{2ab+2ac+2bc}>=\frac{ab+ac+bc+2ab+2ac+2bc}{2ab+2ac+2bc}\)

\(=\frac{3ab+3ac+3bc}{2ab+2ac+2bc}=\frac{3\left(ab+ac+bc\right)}{2\left(ab+ac+bc\right)}=\frac{3}{2}\)(2)

từ (1) và (2)\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>=\frac{3}{2}\)

dấu = xảy ra khi a=b=c