Từ \(\frac{ab}{a+b}=\frac{bc}{b+c}\Leftrightarrow\frac{abc}{ac+bc}=\frac{abc}{ab+ac}\Leftrightarrow bc=ab\Rightarrow a=c\)(1)

Tương tựi ta cũng có : \(\hept{\begin{cases}a=b\\b=c\end{cases}}\)(2)

Từ (1);(2) \(\Rightarrow a=b=c\)Thay vào M ta được :\(M=\frac{a.a+a.a+a.a}{a^2+b^2+c^2}=1\)

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

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

\(\Rightarrow\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}=b=c=a\)

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

chỉ có 1 số 6 thôi,không phải là 66 đâu

ĐKXĐ: \(abc\ne0\)

Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)

\(\Rightarrow x^3+y^3+z^3+\frac{3}{4}=0\Leftrightarrow3xyz+\frac{3}{4}=0\)

\(\Leftrightarrow xyz=-\frac{1}{4}\Leftrightarrow\frac{1}{xyz}=-4\Leftrightarrow abc=-4\)

\(\Rightarrow ab=\frac{-4}{c}\Rightarrow c=Ư\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)

\(c=-4\Rightarrow ab=1\Rightarrow\left(a;b\right)=\left(1;1\right);\left(-1;-1\right)\)

\(c=-2\Rightarrow ab=2\Rightarrow\left(a;b\right)=\left(1;2\right);\left(2;1\right);\left(-2;-1\right);\left(-1;-2\right)\)

....

Tương tự

Ta có \(\left(a-1\right)^2\left(a^2+a+1\right)\ge0\)\(\Leftrightarrow\left(a^2-2a+1\right)\left(a^2+a+1\right)\ge0\)

\(\Leftrightarrow a^4-a^3-a+1\ge0\)

\(\Leftrightarrow a^4-a^3+1\ge a\)

\(\Leftrightarrow a^4-a^3+ab+2\ge a+ab+1\)

\(\Rightarrow\frac{1}{\sqrt{a^4-a^3+ab+2}}\le\frac{1}{\sqrt{ab+a+1}}\)

Tương tự \(\frac{1}{\sqrt{b^4-b^3+bc+2}}\le\frac{1}{\sqrt{bc+b+1}}\)

             \(\frac{1}{\sqrt{c^4-c^3+ca+2}}\le\frac{1}{\sqrt{ca+c+1}}\)

Cộng từng vế các bđt trên ta được

\(VT\le\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\)

Áp dụng bđt Bunhiacopski ta có

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

Dấu "=" xảy ra khi a=b=c=1

Đoán xem

Ta có:

\(M=\frac{19a+3}{1+b^2}+\frac{19b+3}{c^2+1}+\frac{19c+3}{a^2+1}\)

\(=19a-\frac{19ab^2-3}{b^2+1}+19b-\frac{19bc^2-3}{c^2+1}+\frac{19ca^2-3}{a^2+1}\)

\(\ge19\left(a+b+c\right)-\frac{19ab^2-3}{2b}-\frac{19bc^2-3}{2c}-\frac{19ca^2-3}{2a}\)

\(=19\left(a+b+c\right)-19\left(\frac{ab}{2}+\frac{bc}{2}+\frac{ca}{2}\right)+\frac{3}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\ge19.3-\frac{19.3}{2}+\frac{3}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{19.3}{2}+\frac{3}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Lại có:

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\ge3\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\ge3\frac{\left(1+1+1\right)^2}{ab+bc+ca}=\frac{3.9}{3}=9\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\)

\(\Rightarrow M\ge\frac{19.3}{2}+\frac{3}{2}.3=33\)

\(\)

Ta có: \(\frac{19a + 3}{1 + b^2}\)

\(=(19a+3)\left(1-\frac{b^2}{1 + b^2}\right)\)

\(=19a+3-\frac{b^2(19a + 3)}{1 + b^2}\)

Theo AM-GM, ta có:

\(1 + b^2 \ge 2b \implies \frac{1}{1 + b^2} \le \frac{1}{2b}\)

=>\(\frac{b^2(19a + 3)}{1 + b^2} \le \frac{b^2(19a + 3)}{2b} = \frac{19ab + 3b}{2}\)

=>\(\frac{19a + 3}{1 + b^2} \ge 19a + 3 - \frac{19ab + 3b}{2} \quad (1)\)

Chứng minh tương tự, ta cũng có:

\(\frac{19b + 3}{1 + c^2} \ge 19b + 3 - \frac{19bc + 3c}{2} \quad (2)\)

\(\frac{19c + 3}{1 + a^2} \ge 19c + 3 - \frac{19ca + 3a}{2} \quad (3)\)

Từ (1),(2),(3) suy ra \(M \ge 19(a + b + c) + 9 - \frac{19(ab + bc + ca) + 3(a + b + c)}{2}\)

=>\(M\ge19(a+b+c)+9-\frac{19\cdot3+3(a+b+c)}{2}\)

=>\(M \ge 19(a + b + c) + 9 - \frac{57}{2} - \frac{3}{2}(a + b + c)\)

=>\(M \ge \frac{35}{2}(a + b + c) - \frac{39}{2}\)

Ta có: \((a + b + c)^2 \ge 3(ab + bc + ca)\)

=>\((a+b+c)^2\ge3\cdot3=9\)

=>a+b+c>=3

=>\(M \ge \frac{35}{2} \cdot 3 - \frac{39}{2} = \frac{105 - 39}{2} = \frac{66}{2} = 33\)

Dấu '=' xảy ra khi a=b=c=1