gọi A là VT

Ta có : \(A=\left[\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)-x^4y^4\right]+\left[\frac{1}{4}\left(x^{16}+y^{16}\right)-2x^2y^2\right]-1\)

Áp dụng BĐT Cô-si,ta có :

\(\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)\ge\frac{1}{2}2\sqrt{\frac{x^{10}}{y^2}.\frac{y^{10}}{x^2}}=x^4y^4\Rightarrow\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)-x^4y^4\ge0\)

\(\frac{x^{16}+y^{16}}{4}\ge\frac{x^8y^8}{2}=\left(\frac{x^8y^8}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)-\frac{3}{2}\ge4\sqrt[4]{\frac{x^8y^8}{16}}-\frac{3}{2}==2x^2y^2-\frac{3}{2}\)

\(\Rightarrow\frac{1}{4}\left(x^{16}+y^{16}\right)-2x^2y^2\ge\frac{-3}{2}\)

Từ đó ta có : \(A\ge0-\frac{3}{2}-1=\frac{-5}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x=y\\x^2y^2=1\end{cases}\Leftrightarrow x=y=\pm1}\)

\(Q=\left[\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)-x^4y^4\right]+\left[\frac{1}{4}\left(x^{16}+y^{16}\right)-2x^2y^2\right]-1\)

 \(\ge\left(\frac{1}{2}2\sqrt{\frac{x^{10}}{y^2}\cdot\frac{y^{10}}{x^2}}-x^4y^4\right)+\left[\frac{2x^8y^8}{4}-2x^2y^2\right]-1\)

\(\ge\frac{x^8y^8}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}-2x^2y^2-\frac{3}{2}-1\ge4\sqrt[4]{\frac{x^8y^8}{2.2.2.2}}-\frac{3}{2}-1=2x^2y^2-2x^2y^2-\frac{5}{2}=-\frac{5}{2}\)

Vậy min Q = -5/2 tại x = y = +-1 

Còn cách đặt ẩn phụ thế này: 

\(Q=\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)+\frac{1}{4}\left(x^{16}+y^{16}\right)-\left(1+x^2y^2\right)^2\ge\frac{1}{2}.2\sqrt{\frac{x^{10}}{y^2}.\frac{y^{10}}{x^2}}+\frac{1}{4}.2\sqrt{x^{16}.y^{16}}-\left(x^4y^4+2x^2y^2+1\right)\)\(=\frac{x^8y^8}{2}-4x^2y^2-2\)

Đặt x²y² = t >= 0. Khi đó:

\(2Q=t^4-4t-2=\left(t^4-2t^2+1\right)+2\left(t^2-2t+1\right)+5=\left(t^2-1\right)^2+2\left(t-1\right)^2+5\ge5\Rightarrow Q\ge\frac{5}{2}\)Xảy ra đẳng thức khi và chỉ khi x = y =+-1

bạn vào câu hỏi tương tự xem bài của Ngô Thị Thu Trang nhé, Mr.Lazy giải rồi đó

http://olm.vn/hoi-dap/question/147426.html

\(A=\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}+\right)+\frac{1}{4}\left(x^{16}+y^{16}\right)-x^4y^4-2x^2y^2-1\)

Áp dụng Côsi

\(\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)\ge\frac{1}{2}.2\sqrt{\frac{x^{10}}{y^2}.\frac{y^{10}}{x^2}}=x^4y^4\)

\(\frac{1}{4}\left(x^{16}+y^{16}+1+1+1+1+1+1\right)\ge\frac{1}{4}.8\sqrt[8]{x^{16}y^{16}}=2x^2y^2\)

\(\Rightarrow A+\frac{6}{4}\ge x^4y^4+2x^2y^2-x^4y^4-2x^2y^2-1=-1\)

\(\Rightarrow A\ge-1-\frac{6}{4}=-\frac{5}{2}\)

Dấu "=" xảy ra khi và chỉ khi \(x^2=y^2=1\)

Vậy GTNN của A là -2,5 khi x² = y² = 1

\(Q\ge\sqrt{\frac{x^{10}y^{10}}{x^2y^2}}+\frac{1}{2}\sqrt{x^{16}y^{16}}-\left(x^2y^2+1\right)^2\)

\(Q\ge\frac{1}{2}\left(xy\right)^8+\left(xy\right)^4-\left(x^2y^2+1\right)^2\)

Đặt \(x^2y^2=a\ge0\Rightarrow Q\ge\frac{1}{2}a^4+a^2-\left(a+1\right)^2\)

\(Q\ge\frac{1}{2}a^4-2a-1=\frac{1}{2}a^4-2a+\frac{3}{2}-\frac{5}{2}\)

\(Q\ge\frac{1}{2}\left(a-1\right)^2\left(a^2+2a+3\right)-\frac{5}{2}\ge-\frac{5}{2}\)

\(Q_{min}=-\frac{5}{2}\) khi \(a=1\) hay \(x^2=y^2=1\)

\(\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)\ge\frac{1}{2}.2\sqrt{\frac{x^{10}}{y^2}.\frac{y^{10}}{x^2}}=x^4y^4\)

\(x^{16}+y^{16}+1+1+1+1+1+1\ge8\sqrt[8]{x^{16}y^{16}}=8x^2y^2\)

\(\Rightarrow A\ge x^4y^4+\frac{1}{4}\left(8x^2y^2-6\right)-\left(x^4y^4+2x^2y^2+1\right)=-\frac{5}{2}\)

Dấu "=" xảy ra khi \(x^2=y^2=1\)

Vậy GTNN của A là -5/2.

Bài này thắng làm  rồi