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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:
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\ge3\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\ge3\frac{\left(1+1+1\right)^2}{ab+bc+ca}=\frac{3.9}{3}=9\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\)
\(\Rightarrow M\ge\frac{19.3}{2}+\frac{3}{2}.3=33\)
\(\)
Đặt \(x=1-a\), \(y=1-b\), \(z=1-c\)
Ta có : \(1+a=\left(1-b\right)+\left(1-c\right)=y+z\)
\(1+b=\left(1-a\right)+\left(1-c\right)=x+z\)
\(1+c=\left(1-a\right)+\left(1-b\right)=x+y\)
Áp dụng bđt Cauchy, ta có : \(A=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}\ge\frac{2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}}{xyz}=\frac{8xyz}{xyz}=8\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z\Leftrightarrow a=b=c=\frac{1}{3}\)
Vậy Min A = 8 \(\Leftrightarrow a=b=c=\frac{1}{3}\)
\(K=\frac{a^2}{c\left(a^2+c^2\right)}+\frac{b^2}{a\left(a^2+b^2\right)}+\frac{c^2}{b\left(b^2+c^2\right)}\left(a,b,c>0\right)\).
Ta có:
\(\frac{a^2}{c\left(a^2+c^2\right)}=\frac{\left(a^2+c^2\right)-c^2}{c\left(a^2+c^2\right)}=\frac{a^2+c^2}{c\left(a^2+c^2\right)}-\frac{c^2}{c\left(a^2+c^2\right)}\)\(=\frac{1}{c}-\frac{c^2}{c\left(a^2+c^2\right)}\).
Vì \(a,c>0\)nên áp dụng bất đẳng thức Cô-si cho 2 số dương, ta được:
\(a^2+c^2\ge2ac\).
\(\Leftrightarrow c\left(a^2+c^2\right)\ge2ac^2\).
\(\Rightarrow\frac{1}{c\left(a^2+c^2\right)}\le\frac{1}{2ac^2}\)
\(\Leftrightarrow\frac{c^2}{c\left(a^2+c^2\right)}\le\frac{c^2}{2ac^2}=\frac{1}{2a}\).
\(\Leftrightarrow-\frac{c^2}{c\left(a^2+c^2\right)}\ge-\frac{1}{2a}\).
\(\Leftrightarrow\frac{1}{c}-\frac{c^2}{c\left(a^2+c^2\right)}\ge\frac{1}{c}-\frac{1}{2a}\)
\(\Leftrightarrow\frac{a^2}{c\left(a^2+c^2\right)}\ge\frac{1}{c}-\frac{1}{2a}\left(1\right)\)
Dấu bằng xảy ra \(\Leftrightarrow a=c>0\) .
Chứng minh tương tự, ta được:
\(\frac{b^2}{a\left(a^2+b^2\right)}\ge\frac{1}{a}-\frac{1}{2b}\left(a,b>0\right)\left(2\right)\)
Dấu bằng xảy ra \(\Leftrightarrow a=b>0\)
Chứng minh tương tự, ta dược:
\(\frac{c^2}{b\left(b^2+c^2\right)}\ge\frac{1}{b}-\frac{1}{2c}\left(b,c>0\right)\left(3\right)\).
Dấu bằng xảy ra \(\Leftrightarrow b=c>0\).
Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta được:
\(\frac{a^2}{c\left(a^2+c^2\right)}+\frac{b^2}{a\left(a^2+b^2\right)}+\frac{c^2}{b\left(b^2+c^2\right)}\ge\)\(\frac{1}{c}-\frac{1}{2a}+\frac{1}{a}-\frac{1}{2b}+\frac{1}{b}-\frac{1}{2c}\).
\(\Leftrightarrow K\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\).
\(\Leftrightarrow K\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\).
\(\Leftrightarrow K\ge\frac{1}{2}\left(\frac{ab+bc+ca}{abc}\right)\).
Mà \(ab+bc+ca=3abc\)(theo đề bài).
Do đó \(K\ge\frac{1}{2}.\frac{3abc}{abc}\).
\(\Leftrightarrow K\ge\frac{3abc}{2abc}\).
\(\Leftrightarrow K\ge\frac{3}{2}\).
Dấu bằng xảy ra.
\(\Leftrightarrow\hept{\begin{cases}a=b=c>0\\ab+bc+ca=3abc\end{cases}}\Leftrightarrow a=b=c=1\).
Vậy \(minK=\frac{3}{2}\Leftrightarrow a=b=c=1\).
\(\frac{a}{9b^2+1}=\frac{a\left(9b^2+1\right)-9ab^2}{9b^2+1}=a-\frac{9ab^2}{9b^2+1}\ge a-\frac{9ab^2}{2\sqrt{9b^2.1}}=\)
\(=a-\frac{9ab^2}{6b}=a-\frac{3ab}{2}\)
Tương tự với các biểu thức còn lại, kết hợp với
\(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2\)
là được đáp án.