\(\left(x-y\right)^2\ge0;\forall xy\Rightarrow x^2+y^2\ge2xy\)

\(\Rightarrow\left(x+y\right)^2\ge4xy\Rightarrow x+y\ge2\sqrt{xy}\)

\(\dfrac{1}{2}=\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{2}{xy}\Rightarrow xy\ge4\Rightarrow x+y\ge2\sqrt{xy}\ge2\sqrt{4}=4\)

\(C_{min}=4\) khi \(x=y=2\)

Hoặc là:

\(\dfrac{1}{2}=\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(\dfrac{4}{x+y}\right)^2=\dfrac{8}{\left(x+y\right)^2}\)

\(\Rightarrow\left(x+y\right)^2\ge16\Rightarrow x+y\ge4\)

\(P=\left(2x+\dfrac{1}{x}\right)^2+9+\left(2y+\dfrac{1}{y}\right)^2+9-18\)

\(P\ge2\sqrt{9\left(2x+\dfrac{1}{x}\right)^2}+2\sqrt{9\left(2y+\dfrac{1}{y}\right)^2}-18\)

\(P\ge12x+12y+\dfrac{6}{x}+\dfrac{6}{y}-18\)

\(P\ge6\left(4x+\dfrac{1}{x}\right)+6\left(4y+\dfrac{1}{y}\right)-12\left(x+y\right)-18\)

\(P\ge6.2\sqrt{\dfrac{4x}{x}}+6.2\sqrt{\dfrac{4y}{y}}-12.1-18=18\)

\(P_{min}=18\) khi \(x=y=\dfrac{1}{2}\)

Áp dụng BĐT AM-GM ta có:

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

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

\(=\frac{\left(1+4\right)^2}{2}=\frac{5^2}{2}=\frac{25}{2}\)

Xảy ra khi \(x=y=\frac{1}{2}\)

Đề bài sai, C không có giá trị nhỏ nhất

Nếu \(\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{1}{2}\) thì có thể tìm được min của C

cần lưu ý 2 bđt sau :(a,b>0) 1/a + 1/b >= 4/(a+b) , (a+b)² >= 4ab (dâu1 "=" khi a=b)

có x+y=1 =>(x+y)²=1=>x²+y²=1-2xy

A=1/1-2xy + 1/2xy + 1/2xy  >= 4/1-2xy+2xy + 2/4xy >= 4+2/(x+y)² >= 6

Dấu "=" khi x=y=1/2

\(x,y,z>0\)

Áp dụng BĐT Caushy cho 3 số ta có:

\(x^3+y^3+z^3\ge3\sqrt[3]{x^3y^3z^3}=3xyz\ge3.1=3\)

\(P=\dfrac{x^3-1}{x^2+y+z}+\dfrac{y^3-1}{x+y^2+z}+\dfrac{z^3-1}{x+y+z^2}\)

\(=\dfrac{\left(x^3-1\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)}+\dfrac{\left(y^3-1\right)^2}{\left(x+y^2+z\right)\left(y^3-1\right)}+\dfrac{\left(z^3-1\right)^2}{\left(x+y+z^2\right)\left(x^3-1\right)}\)

Áp dụng BĐT Caushy-Schwarz ta có:

\(P\ge\dfrac{\left(x^3+y^3+z^3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}\)

\(\ge\dfrac{\left(3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}=0\)

\(P=0\Leftrightarrow x=y=z=1\)

Vậy \(P_{min}=0\)

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

Ta có bđt sau \(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\) tự chứng mình nha 

Áp dụng \(a=x,b=y,c=1\)

Ta có : \(B=\frac{\left(x+y+1\right)^2}{xy+x+y}\ge3\)

Ta có : \(A=\frac{1}{B}+B=\frac{1}{B}+\frac{B}{9}+\frac{8B}{9}\ge\frac{2}{3}+\frac{8}{3}=\frac{10}{3}\)

Dấu " = " xảy ra khi \(x=y=1\)