Question

[Lâm Ngọc]

cho x,y,z,>0 thỏa mãn xy+yz+xz=3 . tìm min P = \(\sqrt{x^6+y^6+1}+\sqrt{x^6+z^6+1}+\sqrt{z^6+y^6+1}\)

Answer 1

[Kiệt Nguyễn]

Ta có: \(x^3+y^3+z^3=\frac{x^4}{x}+\frac{y^4}{y}+\frac{z^4}{z}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x+y+z}\ge\frac{\left(\frac{\left(x+y+z\right)^2}{3}\right)^2}{x+y+z}\)\(=\frac{\frac{\left(x+y+z\right)^4}{9}}{x+y+z}=\frac{\left(x+y+z\right)^3}{9}\)(Bunhiacopski)

\(\Rightarrow\left(x^3+y^3+z^3\right)^2\ge\frac{\left(x+y+z\right)^6}{81}\ge\frac{\left[3\left(xy+yz+zx\right)\right]^3}{81}=9\)

Áp dụng bất đẳng thức Min - cốp - xki, ta được: \(P=\sqrt{x^6+y^6+1}+\sqrt{y^6+z^6+1}+\sqrt{z^6+x^6+1}\)\(\ge\sqrt{\left(x^3+y^3+z^3\right)^2+\left(x^3+y^3+z^3\right)^2+\left(1+1+1\right)^2}\ge3\sqrt{3}\)

Đẳng thức xảy ra khi x = y = z = 1

Answer 2

[The Angry]

Với ĐK : xy + yz + xz = 3 và Min P = \(\sqrt{x^6+y^6+1}+\sqrt{x^6+z^6+1}+\sqrt{z^6+y^6+1}\) , ta có ĐK :

xy + yz = 3 - xz => xy ; yz ; xz < 3 ; xy = yz = xz = 2 => 2 + 2 + 2 = 3 ( \(\varnothing\)) .

Suy ra : xy = yz = xz = 1 => x = y = z = 1 vì xy + yz + xz = 1 + 1 + 1 = 3.

Từ đó , có tiếp : \(\sqrt{x^6+y^6+1}+\sqrt{x^6+z^6+1}+\sqrt{z^6+y^6+1}\)

\(=\sqrt{1^6+1^6+1}+\sqrt{1^6+1^6+1}\sqrt{1^6+1^6+1}\)

\(=\sqrt{1+1+1}+\sqrt{1+1+1}+\sqrt{1+1+1}\)

\(=3\sqrt{3}\)

=> \(minP=3\sqrt{3}\)