Đặt \(T=\left(a+b\right)\left(b+c\right)\left(c+a\right)>0\)

\(BDT\Leftrightarrow\dfrac{a^2+bc}{b+c}+\dfrac{b^2+ca}{c+a}+\dfrac{c^2+ab}{a+b}\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^2+bc}{b+c}-a+\dfrac{b^2+ca}{c+a}-b+\dfrac{c^2+ab}{a+b}-c\ge0\)

\(\Leftrightarrow\dfrac{a^2+bc-ab-ac}{b+c}+\dfrac{b^2+ac-ab-bc}{a+c}+\dfrac{c^2+ab-ac-bc}{a+b}\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)}{b+c}+\dfrac{\left(b-a\right)\left(b-c\right)}{a+c}+\dfrac{\left(c-a\right)\left(c-b\right)}{a+b}\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)\left(a^2-c^2\right)+\left(b^2-a^2\right)\left(b^2-c^2\right)+\left(c^2-a^2\right)\left(c^2-b^2\right)}{T}\ge0\)

\(\Leftrightarrow\dfrac{a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2}{T}\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2}{2T}\ge0\)

Xảy ra khi \(a=b=c\)

\(BĐT\Leftrightarrow\sum\left(\dfrac{1}{a}-\dfrac{b+c}{a^2+bc}\right)\ge0\)

\(\Leftrightarrow\sum\dfrac{\left(a-b\right)\left(a-c\right)}{a\left(a^2+bc\right)}\ge0\)

Giả sử \(a\ge b\ge c\)thì

\(\dfrac{\left(a-b\right)\left(a-c\right)}{a\left(a^2+bc\right)}\ge0\).vậy nên chỉ cần chứng minh

\(\dfrac{\left(b-c\right)\left(b-a\right)}{b\left(b^2+ac\right)}+\dfrac{\left(c-a\right)\left(c-b\right)}{c\left(c^2+ab\right)}\ge0\)

\(\Leftrightarrow\left(b-c\right)\left[\dfrac{b-a}{b\left(b^2+ac\right)}+\dfrac{a-c}{c\left(c^2+ab\right)}\right]\ge0\)

\(\Leftrightarrow\left(b-c\right)\left[\left(b-a\right)\left(c^3+abc\right)+\left(a-c\right)\left(b^3+abc\right)\right]\ge0\)

\(\Leftrightarrow\left(b-c\right)^2\left(b+c\right)\left(ab+ac-bc\right)\ge0\)( đúng vì \(a\ge b\ge c\))

Vậy BĐT được chứng minh.

Dấu = xảy ra khi a=b=c

Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{b+c}{a^2+bc}=\frac{(b+c)^2}{(b+c)(a^2+bc)}=\frac{(b+c)^2}{b(a^2+c^2)+c(a^2+b^2)}\leq \frac{c^2}{b(a^2+c^2)}+\frac{b^2}{c(a^2+b^2)}\)

Tương tự với các phân thức còn lại:

$\frac{c+a}{b^2+ca}\leq \frac{c^2}{b(a^2+c^2)}+\frac{a^2}{c(a^2+b^2)}$

$\frac{a+b}{c^2+ab}\leq \frac{a^2}{b(a^2+c^2)}+\frac{b^2}{c(a^2+b^2)}$

Cộng theo vế và thu gọn suy ra:

$\text{VT}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ (đpcm)

Dấu "=" xảy ra khi a=b=c

Áp dụng BĐT Cauchy ta có

\(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge a\)

\(\dfrac{b^2}{a+c}+\dfrac{a+c}{4}\ge b\)

\(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}+\dfrac{a+b+c}{2}\ge a+b+c\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)

Dấu bằng xảy ra khi a=b=c

Làm tắt vài chỗ thông cảm

Câu b,

Ta có BĐT Cauchy \(a^2+b^2\ge2ab\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\)

\(\Rightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\)

\(\Rightarrow\dfrac{ab}{a+b}\le\dfrac{\left(a+b\right)^2}{4\left(a+b\right)}=\dfrac{a+b}{4}\)

Tương tự \(\dfrac{bc}{b+c}\le\dfrac{b+c}{4}\)

\(\dfrac{ac}{a+c}\le\dfrac{a+c}{4}\)

Cộng theo vế ta đc \(VT\le\dfrac{2\left(a+b+c\right)}{4}=\dfrac{a+b+c}{2}\)

Dấu bằng xảy ra khi a=b=c

Do \(a;b;c\in\left[0;1\right]\Rightarrow\left(1-a\right)\left(1-c\right)\ge0\)

\(\Leftrightarrow ac+1\ge a+c\)

\(\Rightarrow1+b+ac\ge a+b+c\Rightarrow\dfrac{1}{1+b+ac}\le\dfrac{1}{a+b+c}\)

Tương tự: \(\dfrac{1}{1+c+ab}\le\dfrac{1}{a+b+c}\) ; \(\dfrac{1}{1+a+bc}\le\dfrac{1}{a+b+c}\)

Cộng vế với vế:

\(\dfrac{1}{1+b+ca}+\dfrac{1}{1+c+ab}+\dfrac{1}{1+a+bc}\le\dfrac{3}{a+b+c}\) (đpcm)

Áp dụng bất đẳng thức AM-GM ta có :

\(ab\le\frac{\left(a+b\right)^2}{4}\)=> \(\frac{ab}{a+b}\le\frac{a+b}{4}\)

Tương tự : \(\frac{bc}{b+c}\le\frac{b+c}{4}\); \(\frac{ca}{c+a}\le\frac{c+a}{4}\)

Cộng vế với vế các bđt trên ta có đpcm

Đẳng thức xảy ra <=> a=b=c

Có     $ab\le \genfrac{}{}{0ex}{}{{\left(a+b\right)}^2}{4}\Rightarrow \genfrac{}{}{0ex}{}{ab}{a+b}\le \genfrac{}{}{0ex}{}{a+b}{4}$.

C/m BĐT : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\)

Áp dụng BĐT Sơ-vác-sơ:

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(1+1+1\right)^2}{x+y+z}\ge\dfrac{9}{x+y+z}\)

Ta có: \(9\dfrac{ab}{a+3b+2c}=\dfrac{9ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\left(1\right)\)

CM tương tự

\(\dfrac{9bc}{b+3c+2a}\le\dfrac{bc}{a+c}+\dfrac{bc}{a+b}+\dfrac{b}{2}\left(2\right)\)

\(\dfrac{9ca}{c+3a+2b}\le\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{c}{2}\left(3\right)\)

Cộng vế (1), (2), (3) => đpcm

\(\dfrac{a+b}{ab+c^2}=\dfrac{\left(a+b\right)^2}{\left(ab+c^2\right)\left(a+b\right)}=\dfrac{\left(a+b\right)^2}{b\left(a^2+c^2\right)+a\left(b^2+c^2\right)}\le\dfrac{a^2}{b\left(a^2+c^2\right)}+\dfrac{b^2}{a\left(b^2+c^2\right)}\)

Tương tự: 

\(\dfrac{b+c}{bc+a^2}\le\dfrac{b^2}{c\left(a^2+b^2\right)}+\dfrac{c^2}{b\left(a^2+c^2\right)}\) ; \(\dfrac{c+a}{ca+b^2}\le\dfrac{c^2}{a\left(b^2+c^2\right)}+\dfrac{a^2}{c\left(a^2+b^2\right)}\)

Cộng vế:

\(VT\le\dfrac{1}{a}\left(\dfrac{b^2}{b^2+c^2}+\dfrac{c^2}{b^2+c^2}\right)+\dfrac{1}{b}\left(\dfrac{a^2}{a^2+c^2}+\dfrac{c^2}{a^2+c^2}\right)+\dfrac{1}{c}\left(\dfrac{a^2}{a^2+b^2}+\dfrac{b^2}{a^2+b^2}\right)=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

Theo hệ quả của bất đẳng thức Cauchy

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow3\ge ab+bc+ca\)

\(\Rightarrow\left\{{}\begin{matrix}3+a^2\ge\left(a+c\right)\left(a+b\right)\\3+b^2\ge\left(a+b\right)\left(b+c\right)\\3+c^2\ge\left(a+c\right)\left(b+c\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{bc}{\sqrt{3+a^2}}\le\dfrac{bc}{\sqrt{\left(a+c\right)\left(a+b\right)}}\\\dfrac{ca}{\sqrt{3+b^2}}\le\dfrac{ca}{\sqrt{\left(a+b\right)\left(b+c\right)}}\\\dfrac{ab}{\sqrt{3+c^2}}\le\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\end{matrix}\right.\)

\(\Rightarrow VT\le\dfrac{bc}{\sqrt{\left(a+c\right)\left(a+b\right)}}+\dfrac{ca}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)

\(\Leftrightarrow VT\le\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\) (1)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}\le\dfrac{\dfrac{bc}{a+c}+\dfrac{bc}{a+b}}{2}\\\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\dfrac{ca}{a+b}+\dfrac{ca}{b+c}}{2}\\\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{\dfrac{ab}{a+c}+\dfrac{ab}{b+c}}{2}\end{matrix}\right.\)

\(\Rightarrow\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{\left(\dfrac{bc}{a+c}+\dfrac{ab}{a+c}\right)+\left(\dfrac{bc}{a+b}+\dfrac{ca}{a+b}\right)+\left(\dfrac{ab}{b+c}+\dfrac{ca}{b+c}\right)}{2}\)

\(\Rightarrow\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{a+b+c}{2}=\dfrac{3}{2}\) (2)

Xét \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

\(\Leftrightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\)

Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức

\(\Rightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)

Theo hệ quả của bất đẳng thức Cauchy

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)

\(\Rightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\ge\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\) (3)

Từ (1) , (2) , (3)

\(\Rightarrow VT\le\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

\(\Leftrightarrow\dfrac{bc}{\sqrt{a^2+3}}+\dfrac{ca}{\sqrt{b^2+3}}+\dfrac{ab}{\sqrt{c^2+3}}\le\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\) (đpcm)

Dấu " = " xảy ra khi \(a=b=c=1\)