với các số thực dương a,b,c áp dụng BDT Cauchi ta có:

\(\frac{a^4b}{a^2+1}=a^2b-\frac{a^2b}{a^2+1}\geq a^2b-\frac{a^2b}{2a}=a^2b-\frac{ab}{2}\)

Chứng minh tương tự ta cũng có:
\(\frac{b^4c}{b^2+1}\ge b^2c-\frac{bc}{2},\frac{c^4a}{c^2+1}\ge c^2a-\frac{ca}{2}\)

ta suy ra:

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

áp dụng bdt Cauchy lần nữa, ta có:

\(a^2b+a^2b+b^2c\ge3ab\sqrt[3]{abc}=3ab\)

tương tự ta có:

\(b^2c+b^2c+c^2a\ge3bc\\ c^2a+c^2a+a^2b\ge3ca\)

Vậy:

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

Dấu bằng xảy ra khi\(a=b=c=1\)

ĐK: \(x\ge-1\)

Ta có: \(2x^2-6x+10-5\left(x-2\right)\sqrt{x+1}=0\)

\(\Leftrightarrow2\left(x^2-4x+4\right)+2\left(x+1\right)-5\left(x-2\right)\sqrt{x+1}=0\)

\(\Leftrightarrow2\left(x-2\right)^2+2\left(x+1\right)-5\left(x-2\right)\sqrt{x+1}=0\)

\(\Leftrightarrow\left[2\left(x-2\right)^2-4\left(x-2\right)\sqrt{x+1}\right]-\left[\left(x-2\right)\sqrt{x+1}-2\left(x+1\right)\right]=0\)

\(\Leftrightarrow2\left(x-2\right)\left(x-2-2\sqrt{x+1}\right)-\left(x-2-2\sqrt{x+1}\right)\sqrt{x+1}=0\)

\(\Leftrightarrow\left(2x-4-\sqrt{x+1}\right)\left(x-2-2\sqrt{x+1}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-4=\sqrt{x+1}\\x-2=2\sqrt{x+1}\end{cases}}\left(x\ge2\right)\)

\(\Leftrightarrow\orbr{\begin{cases}4x^2-16x+16=x+1\\x^2-4x+4=4x+4\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x^2-17x+15=0\\x^2-8x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)\left(4x-5\right)=0\\x\left(x-8\right)=0\end{cases}}\Rightarrow x\in\left\{0;\frac{5}{4};3;8\right\}\)

Mà \(x\ge2\) => \(\orbr{\begin{cases}x=3\\x=8\end{cases}\left(tm\right)}\)