Áp dụng liên tục BĐT \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\) kết hợp \(ab\le\frac{\left(a+b\right)^2}{4}\)

\(P=\left(x^4\right)^2+\left(y^4\right)^2+\frac{3}{xy}\ge\frac{\left(x^4+y^4\right)^2}{2}+\frac{3}{\frac{\left(x+y\right)^2}{4}}\ge\frac{\left(\frac{\left(x^2+y^2\right)^2}{2}\right)^2}{2}+\frac{12}{\left(x+y\right)^2}=\frac{\left(x^2+y^2\right)^4}{8}+3\)

\(P\ge\frac{\left(\frac{\left(x+y\right)^2}{2}\right)^4}{8}+3=\frac{\left(x+y\right)^8}{128}+3=5\)

\(\Rightarrow P_{min}=5\) khi \(x=y=1\)

Ta có

3P + 8 = (2a² + 2b²) + (a ² + 4) + (b² + 4)

\(\ge4ab+4a+4b\)= 32

=> P\(\ge\)8

Đạt được khi a = b = 2

\(\sqrt{x^3+8}=\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}\le\frac{x^2-x+6}{2}\)

=>\(\frac{x^2}{\sqrt{x^3+8}}\ge\frac{2x^2}{x^2-x+6}\)

=>A\(\ge\frac{2\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\)

mà \(\left(x+y+z\right)^2\ge3xy+3yz+3zx=9\)

=>\(x+y+z\ge3\)

Xét TS-MS= 2\(4\left(xy+yz+zx\right)+x+y+z-18\ge12+6-18=0\)

=>TS/MS \(\ge1\)

=>A\(\ge1\)

Dấu = khi x=y=z=1

bn có cách giải chưa

bày mk vs

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Lời giải:

Áp dụng BĐT AM-GM:

$x^2+2^2\geq 4x$

$y^2+2^2\geq 4y$

$2(x^2+y^2)\geq 4xy$

$\Rightarrow 3(x^2+y^2)+8\geq 4(x+y+xy)=32$

$\Rightarrow x^2+y^2\geq 8$

Vậy $P_{\min}=8$ khi $x=y=2$

Ta có: \(P=\frac{x^2+y^2}{x-y}=\frac{\left(x-y\right)^2+2xy}{x-y}=\left(x-y\right)+\frac{2xy}{x-y}\)

\(=x-y+\frac{16}{x-y}\ge2.4=8\)

Đặt \(t=x^2+y^2\) thì ta có : 

\(P^2=\frac{\left(x^2+y^2\right)^2}{\left(x-y\right)^2}=\frac{t^2}{t-16}=\frac{1}{\frac{t-16}{t^2}}=\frac{1}{-\frac{16}{t^2}+\frac{1}{t}}=\frac{1}{-16\left(\frac{1}{t}-\frac{1}{32}\right)^2+\frac{1}{64}}\ge\frac{1}{\frac{1}{64}}=64\)

\(\Rightarrow P\ge8\). Đẳng thức xảy ra khi \(\hept{\begin{cases}x^2+y^2=32\\xy=8\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=2+2\sqrt{2}\\y=-2+2\sqrt{3}\end{cases}}\)

Bài 1: Sử dụng phép thế

Có x - y = 2 => x = 2 + y

Thay x = 2 + y vào các biểu thức cần tính

Bài 2:

\(P=9-2\left|x-3\right|\le9\) dấu bằng <=> x = 3

\(Q=\left|x-2\right|+\left|x-8\right|=\left|x-2\right|+\left|8-x\right|\ge\left|x-2+8-x\right|=6\) dấu bằng <=> \(\left(x-2\right)\left(8-x\right)\ge0\)