Xét \(\frac{a+3}{3a+bc}+\frac{b+3}{3b+ca}+\frac{c+3}{3c+ab}\)

\(\Leftrightarrow\frac{2a+b+c}{\left(a+b+c\right)a+bc}+\frac{a+2b+c}{\left(a+b+c\right)b+ca}+\frac{a+b+2c}{\left(a+b+c\right)c+ab}\)

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

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

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

Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm 

\(\Rightarrow\hept{\begin{cases}\left(a+b\right)\left(a+c\right)\le\left(\frac{2a+b+c}{2}\right)^2=\frac{\left(2a+b+c\right)^2}{4}\\\left(b+a\right)\left(b+c\right)\le\left(\frac{a+2b+c}{2}\right)^2=\frac{\left(a+2b+c\right)^2}{4}\\\left(a+c\right)\left(b+c\right)\le\left(\frac{a+b+2c}{2}\right)^2=\frac{\left(a+b+2c\right)^2}{4}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{2a+b+c}{\left(a+b\right)\left(a+c\right)}\ge\frac{4\left(2a+b+c\right)}{\left(2a+b+c\right)^2}=\frac{4}{2a+b+c}\\\frac{a+2b+c}{\left(b+a\right)\left(b+c\right)}\ge\frac{4\left(a+2b+c\right)}{\left(a+2b+c\right)^2}=\frac{4}{a+2b+c}\\\frac{a+b+2c}{\left(a+c\right)\left(b+c\right)}\ge\frac{4\left(a+b+2c\right)}{\left(a+b+2c\right)^2}=\frac{4}{a+b+2c}\end{cases}}\)

\(\Rightarrow VT\ge\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)

Xét \(\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)

Áp dụng bất đẳng thức cộng mẫu số 

\(\Rightarrow\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\ge\frac{\left(2+2+2\right)^2}{2a+b+c+a+2b+c+a+b+2c}\)

\(=\frac{36}{4\left(a+b+c\right)}=\frac{36}{12}=3\)

Mà \(VT\ge\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)

\(\Rightarrow VT\ge3\)

\(\Leftrightarrow\frac{a+3}{3a+bc}+\frac{b+3}{3b+ca}+\frac{c+3}{3c+ab}\ge3\left(đpcm\right)\)

Chúc bạn học tốt !!!

\(VP=\frac{6}{\sqrt{\left(3a+bc\right)\left(3b+ca\right)\left(3c+ab\right)}}\)

\(=\frac{6}{\sqrt{\left[\left(a+b+c\right)a+bc\right]\left[\left(a+b+c\right)b+ca\right]\left[\left(a+b+c\right)c+ab\right]}}\)

\(=\frac{6}{\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+1\right)^2}}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)

\(VT=\frac{1}{3a+bc}+\frac{1}{3b+ca}+\frac{1}{3c+ab}\)

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

\(=\frac{\left(b+c\right)+\left(a+c\right)+\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)

Vậy VT = VP, đẳng thức được chứng minh

Áp dụng BĐT Bunhiacopxki ta có:

\(\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

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

Dấu " = " xảy ra <=> a=b=c=1

Có: \(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2\Leftrightarrow a+b+c\ge3\)( bạn tự c/m nhé )

Dấu " = " xảy ra <=> a=b=c

Áp dụng BĐT Cauchy-schwarz ta có:

\(\frac{a^4}{b+3c}+\frac{b^4}{c+3a}+\frac{c^4}{a+3b}\ge\frac{\left(a^2+b^2+c^2\right)^2}{4\left(a+b+c\right)}\ge\frac{\left[\frac{\left(a+b+c\right)^2}{3}\right]^2}{4\left(a+b+c\right)}=\frac{\left(a+b+c\right)^3}{36}\ge\frac{27}{36}=\frac{3}{4}\)

Dấu " = " xảy ra <=> a=b=c=1 ( bạn tự giải rõ ra nhé )

Ta có:
\(\frac{a^3b}{a^3+b^3}-\frac{ab^3}{a^3+b^3}=\frac{ab\left(a^2-b^2\right)}{a^3+b^3}=\frac{ab\left(a-b\right)}{a^2-ab+b^2}=\frac{a-b}{\frac{a}{b}+\frac{b}{a}-1}\ge\frac{a-b}{\frac{a}{b}+\frac{a}{a}-1}=\frac{b\left(a-b\right)}{a}\)
\(\frac{b^3c}{b^3+c^3}-\frac{bc^3}{b^3+c^3}=\frac{bc\left(b^2-c^2\right)}{b^3+c^3}=\frac{bc\left(b-c\right)}{b^2-bc+c^2}=\frac{b-c}{\frac{b}{c}+\frac{c}{b}-1}\ge\frac{b-c}{\frac{a}{c}+\frac{b}{b}-1}=\frac{c\left(b-c\right)}{a}\)
\(\frac{c^3a}{c^3+a^3}-\frac{ca^3}{c^3+a^3}=\frac{ca\left(c^2-a^2\right)}{c^3+a^3}=\frac{ca\left(c-a\right)}{c^2-ca+a^2}=\frac{c-a}{\frac{c}{a}+\frac{a}{c}-1}\ge\frac{c-a}{\frac{a}{c}+\frac{a}{a}-1}=\frac{c\left(c-a\right)}{a}\)
\(\Rightarrow\frac{a^3b}{a^3+b^3}-\frac{ab^3}{a^3+b^3}+\frac{b^3c}{b^3+c^3}-\frac{bc^3}{b^3+c^3}+\frac{c^3a}{c^3+a^3}-\frac{ca^3}{c^3+a^3}\ge\frac{b\left(a-b\right)+c\left(c-a\right)+c\left(b-c\right)}{a}=\frac{ab-b^2-ac+bc}{a}=\frac{\left(a-b\right)\left(b-c\right)}{a}\ge0\)
\(\Leftrightarrow\frac{a^3b}{a^3+b^3}+\frac{b^3c}{b^3+c^3}+\frac{c^3a}{c^3+a^3}\ge\frac{ab^3}{a^3+b^3}+\frac{bc^3}{b^3+c^3}+\frac{ca^3}{c^3+a^3}\left(đpcm\right)\)

Áp dụng BĐT Bunhiacopxki ta có :

\(\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

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

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

Ta có : \(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2\Leftrightarrow a+b+c\ge3\) ( tự chứng minh ạ )

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

Áp dụng BĐT Cachy Schwarz ta có :

\(\frac{a^4}{b+3c}+\frac{b^4}{c+3a}+\frac{c^4}{a+3b}\ge\frac{\left(a^2+b^2+c^2\right)^2}{4\left(a+b+c\right)}\) \(\ge\frac{\left[\frac{\left(a+b+c\right)}{3}\right]^2}{4\left(a+b+c\right)}=\frac{\left(a+b+c\right)^3}{36}\)

\(\ge\frac{27}{36}=\frac{3}{4}\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\) ( bạn tự giải rõ ạ )

Thay \(a+b+c=3\) ta được:

\(VT=\frac{1}{a\left(a+b+c\right)+bc}+\frac{1}{b\left(a+b+c\right)+ca}+\frac{1}{c\left(a+b+c\right)+ab}\)

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

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

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

\(=\frac{b+c+a+c+a+b}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=\frac{2\left(a+b+c\right)}{\sqrt{\left[\left(a+b\right)\left(a+c\right)\right].\left[\left(a+b\right)\left(b+c\right)\right].\left[\left(a+c\right)\left(b+c\right)\right]}}\)

\(=\frac{6}{\sqrt{\left(3a+bc\right)\left(3b+ca\right)\left(3c+ab\right)}}=VP\)  (Do \(a+b+c=3\))

=> ĐPCM.