ta có 

\(x^4+1\ge2x^2\text{ nên }x^4+x^2+1\ge3x^2\)

Nên \(A=\frac{x^2}{x^4+x^2+1}\le\frac{1}{3}\)Vậy GTLN A =1/3

GTNN cua A thì do \(\hept{\begin{cases}x^2\ge0\\x^4+x^2+1>0\end{cases}\Rightarrow A\ge0}\) nên GTNN A=0

Ta có: 

\(S=\left(xy+\frac{1}{xy}\right)^2=x^2y^2+\frac{1}{x^2y^2}+2\)

\(=\left(x^2y^2+\frac{1}{256x^2y^2}\right)+\frac{255}{256x^2y^2}+2\)

\(\ge2\sqrt{x^2y^2\cdot\frac{1}{256x^2y^2}}+\frac{255}{256\left[\frac{\left(x+y\right)^2}{4}\right]^2}+2\)

\(=2\cdot\frac{1}{16}+\frac{255}{256\cdot\frac{1}{16}}+2=\frac{289}{16}\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)

Ta có:

\(3x^2-6x+10-2\sqrt{x^3+1}-\sqrt{\frac{x^4+16}{2}}=0\)

Ta có:

\(VT=\left(x-2\right)^2+2x^2-2x+6-2\sqrt{\left(x+1\right)\left(x^2-x+1\right)}-\sqrt{\frac{x^4+16}{2}}\)

\(\ge2x^2-2x+6-\left(x+1+x^2-x+1\right)-\sqrt{\frac{x^4+16}{2}}\)

\(\ge x^2-2x+4-\sqrt{\frac{x^4+16}{2}}\)

Ta chứng minh:

\(x^2-2x+4-\sqrt{\frac{x^4+16}{2}}\ge0\)

\(\Leftrightarrow x^2-2x+4\ge\sqrt{\frac{x^4+16}{2}}\)

\(\Leftrightarrow\left(x^2-2x+4\right)^2\ge\frac{x^4+16}{2}\)

\(\Leftrightarrow\left(x-2\right)^4\ge0\)

Vậy \(VT\ge VP\)dấu = xảy ra khi \(x=2\)