Với x dương, ta có đánh giá:

\(\dfrac{x}{1+x^2}\le\dfrac{36x+3}{50}\)

Thật vậy, BĐT tương đương:

\(\left(x^2+1\right)\left(36x+3\right)\ge50x\)

\(\Leftrightarrow36x^3+3x^2-14x+3\ge0\)

\(\Leftrightarrow\left(3x-1\right)^2\left(4x+3\right)\ge0\) (luôn đúng)

Áp dụng:

\(\dfrac{10a}{1+a^2}+\dfrac{10b}{1+b^2}+\dfrac{10c}{1+c^2}\le10.\dfrac{36\left(a+b+c\right)+9}{50}=9\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Mẫu số to quá nên ko nghĩ ra cách giải đẹp mắt:

Dự đoán dấu "=" xảy ra tại \(a=b=c=1\), ta cần c/m: \(A\le\dfrac{3}{16}\)

Do \(\sum\dfrac{a+1}{a^2+1+10a+20}\le\sum\dfrac{a+1}{2a+10a+20}=\sum\dfrac{a+1}{12a+20}\)

Nên ta chỉ cần chứng minh: \(\sum\dfrac{a+1}{3a+5}\le\dfrac{3}{4}\Leftrightarrow\sum\left(\dfrac{3a+3}{3a+5}-1\right)\le\dfrac{9}{4}-3\)

\(\Leftrightarrow\sum\dfrac{1}{3a+5}\ge\dfrac{3}{8}\Leftrightarrow\dfrac{3\left(ab+bc+ca\right)+10\left(a+b+c\right)+25}{\left(3a+5\right)\left(3b+5\right)\left(3c+5\right)}\ge\dfrac{1}{8}\) (quy đồng)

\(\Leftrightarrow\dfrac{4\left(a+b+c\right)+3\left(ab+bc+ca+2\left(a+b+c\right)\right)+25}{27abc+45\left(ab+bc+ca+2\left(a+b+c\right)\right)-15\left(a+b+c\right)+125}\ge\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4\left(a+b+c\right)+52}{27abc-15\left(a+b+c\right)+530}\ge\dfrac{1}{8}\)

\(\Leftrightarrow47\left(a+b+c\right)\ge27abc+114\)

Điều này đúng do:

\(9=2\left(a+b+c\right)+ab+bc+ca\le2\left(a+b+c\right)+\dfrac{1}{3}\left(a+b+c\right)^2\)

\(\Rightarrow\left(a+b+c-3\right)\left(a+b+c+9\right)\ge0\)

\(\Rightarrow a+b+c\ge3\)

Và: \(9=a+b+c+a+b+c+ab+bc+ca\ge9\sqrt[9]{a^4b^4c^4}\)

\(\Rightarrow abc\le1\)

\(\Rightarrow\left\{{}\begin{matrix}47\left(a+b+c\right)\ge141\\27abc+114\le27+114=141\end{matrix}\right.\) (đpcm)

Ta có : \(\frac{10a+b}{a+b}=\frac{10b+c}{b+c}\Leftrightarrow10ab+10ac+b^2+bc=10ab+10b^2+ca+cb\)

\(\Leftrightarrow\)9ac=9b² \(\Leftrightarrow\)\(\frac{a}{b}=\frac{b}{c}\)