DO \(a+b+c=0\)

=>\(a^3+b^3+c^3=3abc\)

 DO \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

=> \(ab+ac+bc=0\)

TA CÓ \(\left(a^3+b^3+c^3\right)^2\)

       = \(a^6+b^6+c^6+2\left(a^3b^3+b^3c^3+a^3c^3\right)=9a^2b^2c^2\)

DO \(ab+ac+bc=0\)

=> \(a^3b^3+b^3c^3+a^3c^3=0\)

=> \(a^6+b^6+c^6=9a^2b^2c^2\)

=> \(\frac{a^6+b^6+c^6}{a^3+b^3+c^3}=\frac{9a^2b^2c^2}{3abc}=3abc\)

Ta có\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) nên ab + bc + ca = 0. Kết hợp với a + b + c = 0 ta được a² + b² + c² = 0.

Sử dụng phân tích: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca) trong điều kiện a + b + c = 0 và a² + b² + c² = 0 ta được:

nên a³ + b³ + c³ = 3abc.   (1)

và a⁶ + b⁶ + c⁶ = 3a²b²c².   (2)

từ (1) và (2) suy ra đpcm.

Áp dụng BĐT Cauchy-Schwarz ta có:

\(P=\frac{1}{\left(a+2\right)+\left(a+2\right)+\left(b+2\right)}+\frac{1}{\left(b+2\right)+\left(b+2\right)+\left(c+2\right)}+\frac{1}{\left(c+2\right)+\left(c+2\right)+\left(a+2\right)}\)

\(\le\frac{1}{9}\left(\frac{2}{a+2}+\frac{1}{b+2}\right)+\frac{1}{9}\left(\frac{2}{b+2}+\frac{1}{c+2}\right)+\frac{1}{9}\left(\frac{2}{c+2}+\frac{1}{a+2}\right)\)

\(=\frac{1}{3}\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)\)

Dễ dàng cm BĐT \(\frac{1}{x+1}+\frac{1}{y+1}\ge\frac{2}{1+\sqrt{xy}}\)

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

\(\le\frac{1}{2}.\frac{3}{1+\sqrt[3]{\frac{abc}{8}}}=\frac{3}{4}\Rightarrow P\le\frac{1}{4}\)

Xảy ra khi \(a=b=c=2\)

À viết ngược dấu BĐT phụ r` :v

\(\frac{1}{1+x}+\frac{1}{1+y}\le\frac{2}{1+\sqrt{xy}}\) mới đúng nhé :v

\(\Leftrightarrow\frac{\left(\sqrt{xy}-1\right)\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(x+1\right)\left(y+1\right)\left(1+\sqrt{xy}\right)}\le0\) 

xy >/ 1 ? 

+ chứng bất đẳng thức phụ: \(\frac{1}{x+y}\le\frac{1}{4x}+\frac{1}{4y}\left(x,y>0\right)\) 

  Với \(x,y>0:\left(x-y\right)^2\ge0\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow x^2+2xy+y^2\ge4xy\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow\frac{x+y}{4xy}\ge\frac{1}{x+y}\Leftrightarrow\frac{1}{x+y}\le\frac{1}{4x}+\frac{1}{4y}\)(đpcm)

 Dấu "=" xảy ra \(\Leftrightarrow x-y=0\Leftrightarrow x=y\)

+ Thay \(a+b+c=6\)vào P , ta được: \(P=\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{c+a}\)

 Áp dụng bđt chứng minh trên , ta được:\(\frac{1}{a+b}\le\frac{1}{4a}+\frac{1}{4b}\Rightarrow\frac{ab}{a+b}\le ab\left(\frac{1}{4a}+\frac{1}{4b}\right)=\frac{a}{4}+\frac{b}{4}\)

 Tương tự như vậy rồi cộng từng vế các bđt , ta được 

\(P\le\frac{a}{4}+\frac{b}{4}+\frac{b}{4}+\frac{c}{4}+\frac{c}{4}+\frac{a}{4}=\frac{a+b+c}{2}=\frac{6}{2}=3\)

Dấu "=" xảy ra\(\Leftrightarrow\hept{\begin{cases}a=b=c\\a+b+c=6\end{cases}}\Leftrightarrow a=b=c=2\)

 Vậy maxP =3\(\Leftrightarrow a=b=c=2\)

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

áp dụng bđt cauchy ta có:

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

\(\Rightarrow M\ge\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\ge3\sqrt[3]{\frac{1}{8abc}}=\frac{3}{2}\)

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

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

Ta có:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Leftrightarrow ab+bc+ca=0\)

Mà \(\left(a+b+c\right)^2=0\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)

\(\Rightarrow a^2+b^2+c^2=0\)

Ta lại có:

\(\frac{a^6+b^6+c^6}{a^3+b^3+c^3}=\frac{\left(a^6+b^6+c^6-3a^2b^2c^2\right)+3a^2b^2c^2}{\left(a^3+b^3+c^3-3abc\right)+3abc}\)

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

\(=\frac{3a^2b^2c^2}{3abc}=abc\)