Ace Legona Bạn mà ko giải được thì còn ai giải đc nữa mà hỏi

a/ \(\frac{4bc-a^2}{bc+2a^2}.\frac{4ab-c^2}{ab+2c^2}.\frac{4ac-b^2}{ac+2b^2}\)

\(=\frac{4bc-\left(b+c\right)^2}{bc+2\left(b+c\right)^2}.\frac{4\left(-b-c\right)b-c^2}{\left(-b-c\right)b+2c^2}.\frac{4\left(-b-c\right)c-b^2}{\left(-b-c\right)c+2b^2}\)

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

\(P=\frac{bc}{2ab+ac}+\frac{ca}{2ab+bc}+\frac{4ab}{bc+ca}\)

Xét \(Q=P+3=\frac{bc}{2ab+ac}+1+\frac{ca}{2ab+bc}+1+\frac{4ab}{bc+ca}+1\)

\(Q=\frac{2ab+ac+bc}{2ab+ac}+\frac{2ab+ac+bc}{2ab+bc}+\frac{4ab+bc+ca}{bc+ca}\)

\(=\left(2ab+ac+bc\right)\left(\frac{1}{2ab+ac}+\frac{1}{2ab+bc}\right)+\frac{4ab+bc+ca}{bc+ca}\)

\(\ge\left(2ab+ac+bc\right)\frac{4}{4ab+ac+bc}+\frac{4ab+bc+ca}{bc+ca}=K\)(Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)với a, b không âm)

\(K=\frac{2\left(4ab+ac+bc\right)+2\left(ac+bc\right)}{4ab+ac+bc}+\frac{2\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}\)\(+\frac{7\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}\)

\(=2+\left[\frac{2\left(ac+bc\right)}{4ab+ac+bc}+\frac{2\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}\right]+\frac{7}{9}+\frac{7}{9}.\frac{4ab}{ac+bc}\)

\(\ge2+2\sqrt{\frac{2\left(ac+bc\right)}{4ab+ac+bc}.\frac{2\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}}+\frac{7}{9}+\frac{7}{9}.\frac{4ab}{ac+bc}\)(Áp dụng BĐT Cô - si cho 2 số không âm)

\(=\frac{37}{9}+\frac{7}{9}.\frac{4ab}{ac+bc}\)

Mặt khác: \(6=2\left(\frac{a}{b}+\frac{b}{a}\right)+c\left(\frac{a}{b^2}+\frac{b}{a^2}\right)=\frac{2\left(a^2+b^2\right)}{ab}+\frac{c\left(a^3+b^3\right)}{a^2b^2}\)

\(=\frac{2\left(a^2+b^2\right)}{ab}+\frac{c\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2b^2}\)\(\ge\frac{2.2ab}{ab}+\frac{c\left(a+b\right)\left(2ab-ab\right)}{a^2b^2}=4+\frac{ac+bc}{ab}\)(theo BĐT \(a^2+b^2\ge2ab\))

\(\Rightarrow\frac{ac+bc}{ab}\le2\Leftrightarrow\frac{ab}{ac+bc}\ge\frac{1}{2}\)

\(\Rightarrow K\ge\frac{37}{9}+\frac{7}{9}.\frac{4ab}{ac+bc}\ge\frac{37}{9}+\frac{7}{9}.\frac{4}{2}=\frac{17}{3}\)

Ta có \(Q=P+3\ge K\ge\frac{17}{3}\Rightarrow P\ge\frac{17}{3}-3=\frac{8}{3}\)

Đẳng thức xảy ra khi \(\hept{\begin{cases}2ab+ac=2ab+bc\\\frac{2\left(ac+bc\right)}{4ab+ac+bc}=\frac{2\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}\\a=b\end{cases}}\)\(\Leftrightarrow a=b=c\)

Từ \(2\left(\frac{a}{b}+\frac{b}{a}\right)+c\left(\frac{a}{b^2}+\frac{b}{a^2}\right)=6\Rightarrow6=\frac{c\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2b^2}+\frac{2\left(a^2+b^2\right)}{ab}\)

ta có \(a^2+b^2\ge2ab\Rightarrow6=\frac{c\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2b^2}+\frac{2\left(a^2+b^2\right)}{ab}\ge\frac{c\left(a+b\right)}{ab}+4\)

\(\Rightarrow0< \frac{c\left(a+b\right)}{ab}\le2\)

Lại có 

\(\frac{bc}{a\left(2b+c\right)}+\frac{ac}{b\left(2a+c\right)}=\frac{\left(bc\right)^2}{abc\left(2b+c\right)}+\frac{\left(ac\right)^2}{abc\left(2a+c\right)}\ge\frac{\left(bc+ac\right)^2}{2abc\left(a+b+c\right)}\)\(=\frac{\left[c\left(a+b\right)\right]^2}{2abc\left(a+b+c\right)}\)

và \(abc\left(a+b+c\right)=ab\cdot bc+bc\cdot ba+ab\cdot ca\le\frac{\left(ab+bc+ca\right)^2}{3}\)

\(\Rightarrow\frac{bc}{a\left(2b+c\right)}+\frac{ac}{b\left(2a+c\right)}\ge\frac{3}{2}\left(\frac{c\left(a+b\right)}{ab+bc+ca}\right)^2=\frac{3}{2}\left(\frac{\frac{c\left(a+b\right)}{ab}}{1+\frac{c\left(a+b\right)}{ab}}\right)^2\)

Đặt \(t=\frac{c\left(a+b\right)}{ab}\Rightarrow P\ge\frac{3t^2}{2\left(1+t\right)^2}+\frac{4}{t}\left(0< t\le2\right)\)

Có \(\frac{3t^2}{2\left(1+t\right)^2}+\frac{4}{t}=\left(\frac{3t^2}{\left(1+t\right)^2}+\frac{4}{t}-\frac{8}{3}\right)+\frac{8}{3}=\frac{-7t^2-8t^2+32t+24}{6t\left(1+t\right)^2}+\frac{8}{3}\)

\(=\frac{\left(t-2\right)\left(-7t^2-22t-12\right)}{6t\left(1+t\right)^2}\ge0\forall t\in(0;2]\)

=> \(\frac{\left(t-2\right)\left(-7t^2-22t-12\right)}{6t\left(1+t\right)^2}+\frac{8}{3}\ge\frac{8}{3}\forall t\in(0;2]\frac{1}{2}\)

Dấu "=" xảy ra <=> t=2 hay a=b=c

Áp dụng bất đẳng thức Cauchy Schwarz thôi bạn

\(S\ge3\frac{1}{\sqrt[6]{\left(a+2b+5c\right)\left(b+2c+5a\right)\left(c+2a+5b\right)}}.\)
\(S\ge\frac{3.4}{\sqrt[6]{\left(a+2b+5c\right)\left(b+2c+5a\right)\left(c+2a+5b\right).16.16.16}}\)

\(S\ge\frac{12}{\frac{a+2b+5c+b+2c+5a+c+2a+5b+16+16+16}{6}}\)

\(S\ge\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=2\)

Áp dụng BĐT AM-GM ta có:

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

\(\ge4+\frac{c\left(a^3+b^3\right)}{a^2b^2}\ge4+\frac{c\left(a+b\right)}{ab}\)\(\Rightarrow\frac{c\left(a+b\right)}{ab}\in\text{(}0;2\text{]}\)

Áp dụng BĐT Cauchy-Schwarz lại có:

\(P\ge\frac{\left(bc+ca\right)^2}{2abc\left(a+b+c\right)}+\frac{4}{\frac{c\left(a+b\right)}{ab}}\)\(\ge\frac{3c^2\left(a+b\right)^2}{2\left(ab+bc+ca\right)}+\frac{4}{\frac{c\left(a+b\right)}{ab}}\)

\(=\frac{\frac{3c^2\left(a+b\right)^2}{a^2b^2}}{2\left(1+\frac{ca}{ab}+\frac{bc}{ab}\right)^2}+\frac{4}{\frac{c\left(a+b\right)}{ab}}\)

\(=\frac{\frac{3c^2\left(a+b\right)^2}{a^2b^2}}{2\left[1+\frac{c\left(a+b\right)}{ab}\right]^2}+\frac{4}{\frac{c\left(a+b\right)}{ab}}\)

Đặt \(x=\frac{c\left(a+b\right)}{ab}\left(x\in\text{(}0;2\text{]}\right)\) khi đó ta có:

\(P\ge\frac{3x^2}{2\left(1+x\right)^2}+\frac{4}{x}\) cần chứng minh \(P\ge\frac{8}{3}\Leftrightarrow\left(x-2\right)\left(7x^2+22x+12\right)\le0\forall x\in\text{(0;2]}\)

Vậy \(Min_P=\frac{8}{3}\) khi a=b=c=2

Chỗ dùng cauchy- schwarz mình không hiểu lắm

\(C=\frac{4ab}{a+2b}+\frac{9ac}{4c+a}+\frac{4bc}{b+c}=\frac{4abc}{ac+2bc}+\frac{9abc}{4bc+ab}+\frac{4abc}{ab+ac}\)

\(\ge\frac{\left(2\sqrt{abc}+3\sqrt{abc}+2\sqrt{abc}\right)^2}{ac+2bc+4bc+ab+ab+ac}=\frac{49abc}{2ac+6bc+2ab}=7\)

Xin bổ sung cách sau, bn có thể tham khảo thêm

:\(GT\Leftrightarrow\frac{2}{c}+\frac{6}{a}+\frac{2}{b}=7\)

Đặt \(\hept{\begin{cases}\frac{1}{c}=x\\\frac{1}{b}=y\\\frac{3}{a}=z\end{cases}}\) Ta có: \(2\left(x+y+z\right)=7\)

Suy ra \(C=\frac{4}{4y+\frac{2z}{3}}+\frac{9}{x+\frac{4z}{3}}+\frac{4}{x+y}\ge\frac{\left(2+3+2\right)^2}{2\left(x+y+z\right)}=7\) (Bdt Cauchy-Schwarz)

Dấu = khi \(\hept{\begin{cases}a=2\\b=c=1\end{cases}}\)

Ta có \(2ab+6bc+2ca=7abc\Leftrightarrow\frac{6}{a}+\frac{2}{b}+\frac{2}{c}=7\)

Ta có:

\(C=\frac{4ab}{a+2b}+\frac{9ac}{a+4c}+\frac{4bc}{b+c}=\frac{2^2}{\frac{1}{b}+\frac{2}{a}}+\frac{3^2}{\frac{1}{c}+\frac{4}{a}}+\frac{2^2}{\frac{1}{c}+\frac{1}{b}}\ge\frac{\left(2+3+2\right)^2}{\frac{6}{b}+\frac{2}{b}+\frac{2}{c}}=\frac{49}{7}=7\)

Vậy \(Min_C=7\Leftrightarrow\hept{\begin{cases}a=2b=2c\\\frac{6}{a}+\frac{2}{b}+\frac{2}{c}=7\end{cases}\Leftrightarrow\hept{\begin{cases}a=2\\b=c=1\end{cases}}}\)