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

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

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

\(=\frac{ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

Cần chứng minh \(\frac{ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{3}{4}\)

\(\Leftrightarrow a^2b+a^2c+ab^2+ac^2+b^2c+bc^2\ge6abc\)

BĐT cuối luôn đúng theo AM-GM

\(\frac{a}{b^2+bc+c^2}+\frac{b}{c^2+ca+a^2}+\frac{c}{a^2+ab+b^2}=\frac{a^2}{ab^2+abc+ac^2}+\frac{b^2}{bc^2+abc+ba^2}+\frac{c^2}{ca^2+abc+cb^2}\)     (1)

Áp dụng BDT Cauchy-Schwarz: \(\left(1\right)\ge\frac{\left(a+b+c\right)^2}{ab^2+ac^2+ba^2+bc^2+ca^2+cb^2+3abc}\)

Lại có: \(ab^2+ac^2+ba^2+bc^2+ca^2+cb^2+3abc=\left(ab+bc+ac\right)\left(a+b+c\right)\)

Thay vào -> dpcm

\(VT=\frac{a^2}{ab^2+abc+ac^2}+\frac{b^2}{c^2b+abc+a^2b}+\frac{c^2}{a^2c+abc+b^2c}\)

Áp dụng BĐT Cauchy dạng phân thức 

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

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

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

\(\Leftrightarrow VT\ge\frac{a+b+c}{ab+bc+ac}\left(đpcm\right)\)

Dấu "=" xảy ra khi a=b=c

Chúc bạn học tốt !!!

Áp dụng BĐT AM-GM ta có:

\(\hept{\begin{cases}\frac{bc}{a}+\frac{ac}{b}\ge2.\sqrt{\frac{bc}{a}.\frac{ac}{b}}=2.c\\\frac{bc}{a}+\frac{ab}{c}\ge2.\sqrt{\frac{bc}{a}.\frac{ab}{c}}=2b\\\frac{ac}{b}+\frac{ab}{c}\ge2.\sqrt{\frac{ac}{b}.\frac{ab}{c}}=2a\end{cases}}\Leftrightarrow2.\left(\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\right)\ge2\left(a+b+c\right)\)

\(\Leftrightarrow\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\ge a+b+c\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)( tự giải rõ ra nhé )

BĐT AM-GM: 

\(a+a_1+a_2+...+a_n\ge n\sqrt[n]{a.a_1.a_2.....a_n}\)

Dấu " = " xảy ra \(\Leftrightarrow a=a_1=a_2=...=a_n\)

\(\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\ge a+b+c\)

\(\Leftrightarrow\frac{abc}{a^2}+\frac{abc}{b^2}+\frac{abc}{c^2}\ge a+b+c\)

\(\Leftrightarrow abc.\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge a+b+c\)

Giải tiếp nhé

Bài làm :

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

\(=\frac{1}{b}\cdot\frac{a}{b^2+a}+\frac{1}{c}\cdot\frac{b}{c^2+b}+\frac{1}{a}\cdot\frac{c}{a^2+c}\)

\(=\frac{1}{b}\cdot\left(1-\frac{b^2}{b^2+a}\right)+\frac{1}{c}\cdot\left(1-\frac{c^2}{c^2+b}\right)+\frac{1}{a}\cdot\left(1-\frac{a^2}{a^2+c}\right)\)

Áp dụng BĐT Cô-si :

\(VT\ge\frac{1}{b}\cdot\left(1-\frac{b^2}{2b\sqrt{a}}\right)+\frac{1}{c}\cdot\left(1-\frac{c^2}{2c\sqrt{b}}\right)+\frac{1}{a}\cdot\left(1-\frac{a^2}{2a\sqrt{c}}\right)\)

\(=\frac{1}{b}\cdot\left(1-\frac{b}{2\sqrt{a}}\right)+\frac{1}{c}\cdot\left(1-\frac{c}{2\sqrt{b}}\right)+\frac{1}{a}\cdot\left(1-\frac{a}{2\sqrt{c}}\right)\)

\(=\frac{1}{b}-\frac{1}{2\sqrt{a}}+\frac{1}{c}-\frac{1}{2\sqrt{b}}+\frac{1}{a}-\frac{1}{2\sqrt{c}}\)

\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{2}\cdot\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\)

Lại áp dụng BĐT Cô-si :

\(\frac{1}{\sqrt{a}}\le\frac{\frac{1}{a}+1}{2};\frac{1}{\sqrt{b}}\le\frac{\frac{1}{b}+1}{2};\frac{1}{\sqrt{c}}\le\frac{\frac{1}{c}+1}{2}\)

Do đó :

\(VT\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{2}\cdot\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3}{2}\)

\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{4}\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{3}{4}\)

\(=\frac{3}{4}\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{3}{4}\ge\frac{3}{4}\cdot\frac{9}{a+b+c}-\frac{3}{4}=\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)

giúp vs

Lê Thị Thục HiềnTrần Thanh PhươngVũ Minh Tuấn

\(VT=\frac{a^2}{ab^2+abc+ac^2}+\frac{b^2}{c^2b+abc+a^2b}+\frac{c^2}{a^2c+abc+b^2c}\)

Áp dụng bđt Cauchy dạng phân thức 

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

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

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

\(\Leftrightarrow VT\ge\frac{a+b+c}{ab+bc+ac}\left(đpcm\right)\)

Dấu " = " xảy ra khi \(a=b=c\)

Chúc bạn học tốt !!!

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