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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
\(\Rightarrow\left\{\begin{matrix}\left(a+b\right)\left(a+c\right)\le\left(\frac{2a+b+c}{2}\right)^2=\frac{\left(2a+b+c\right)^2}{4}\\\left(b+a\right)\left(b+c\right)\le\left(\frac{a+2b+c}{2}\right)^2=\frac{\left(a+2b+c\right)^2}{4}\\\left(a+c\right)\left(b+c\right)\le\left(\frac{a+b+2c}{2}\right)^2=\frac{\left(a+b+2c\right)^2}{4}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{2a+b+c}{\left(a+b\right)\left(a+c\right)}\ge\frac{4\left(2a+b+c\right)}{\left(2a+b+c\right)^2}=\frac{4}{2a+b+c}\\\frac{a+2b+c}{\left(b+a\right)\left(b+c\right)}\ge\frac{4\left(a+2b+c\right)}{\left(a+2b+c\right)^2}=\frac{4}{a+2b+c}\\\frac{a+b+2c}{\left(a+c\right)\left(b+c\right)}\ge\frac{4\left(a+b+2c\right)}{\left(a+b+2c\right)^2}=\frac{4}{a+b+2c}\end{matrix}\right.\)
\(\Rightarrow VT\ge\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)
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 \(VT\ge\frac{4}{2a+b+c}+\frac{4}{a+2b+c}+\frac{4}{a+b+2c}\)
\(\Rightarrow VT\ge3\)
\(\Leftrightarrow\frac{a+3}{3a+bc}+\frac{b+3}{3b+ca}+\frac{c+3}{3c+ab}\ge3\) ( đpcm )
Ta có:
\(3a+bc=(a+b+c)a+bc=(a+c)(a+b)\)
\(\Rightarrow \sum \frac{a+3}{3a+bc}\)\(= \sum \frac{(a+c)+(a+b)}{(a+c)(a+b)}=2 \sum \frac{1}{a+b}\geq 2.\frac{9}{2(a+b+c)}=3\)
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 !!!
Áp dụng BĐT C-S dạng ENgel ta có:
$$\frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a} \ge \frac{3}{3+abc} $$
$$\frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a} \ge \frac{9}{4(a+b+c)} $$
Ta chứng minh $$ \frac{9}{4(a+b+c)} \ge \frac{3}{3+abc} $ hay $9+3abc \ge 4(a+b+c) $$
Đặt $ a= 1-x, b=1-y, c=1-z $ rồi xài AM-GM
Tử là mũ 2 thật hả bạn. Mũ 3 thì giải được còn mũ 2 thì vẫn chưa nghĩ ra
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{c+a}\geq \frac{9}{b+c+c+a+c+a}=\frac{9}{3c+2a+b}\)
\(\frac{1}{a+c}+\frac{1}{a+b}+\frac{1}{a+b}\geq \frac{9}{a+c+a+b+a+b}=\frac{9}{3a+2b+c}\)
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{b+c}\geq \frac{9}{a+b+b+c+b+c}=\frac{9}{3b+2c+a}\)
Cộng theo vế rồi rút gọn ta thu được
\(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\geq 3\left(\frac{1}{3a+2b+c}+\frac{1}{3b+2c+a}+\frac{1}{3c+2a+b}\right)\) (đpcm)
Dấu bằng xảy ra khi $a=b=c$
Áp dụng Cauchy Schwarz dạng Engel ta có :
\(\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=\frac{a+b+c}{5}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)