Question

[Ngô Thanh Huyền]

cho x,y,z là 3 số dương và x+y+y \(\le\)1 . CMR : \(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\ge82\)

Answer 1

[G-Dragon]

Ta có :

\(VT\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\) ( Sử dụng phương pháp véctơ )

Do đó :

\(VT^2=\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)\(=81\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)\(-80\left(x+y+z\right)^2\ge18\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-80\left(x+y+z\right)^2\)\(\ge162-80=82\)

\(\Rightarrow VT\ge\sqrt{82}\)

Đẳng thức xảy ra khi x = y = z = \(\frac{1}{3}\)

Answer 2

[Lê Thị Thục Hiền]

Cách khác

Áp dụng bđt bunhiacopski có:

\(\left(1.x+9.\frac{1}{x}\right)^2\le\left(1^2+9^2\right)\left(x^2+\frac{1}{x^2}\right)\)

=> \(\sqrt{x^2+\frac{1}{x^2}}\ge\frac{\left(x+\frac{9}{x}\right)}{\sqrt{82}}\)

CM tương tự: \(\sqrt{y^2+\frac{1}{y^2}}\ge\frac{\left(y+\frac{9}{y}\right)}{\sqrt{82}}\)

\(\sqrt{z^2+\frac{1}{z^2}}\ge\frac{\left(z+\frac{9}{z}\right)}{\sqrt{82}}\)

Cộng vế với vế =>A= \(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\ge\frac{\left(x+y+z+\frac{9}{x}+\frac{9}{y}+\frac{9}{z}\right)}{\sqrt{82}}\)

Áp dụng svac-xơ vào VP có A \(\ge\frac{\left(x+y+z+\frac{81}{x+y+z}\right)}{\sqrt{82}}=\frac{\left(x+y+z+\frac{1}{x+y+z}+\frac{80}{x+y+z}\right)}{\sqrt{82}}\ge\frac{\left(2+80\right)}{\sqrt{82}}\)

<=> \(A\ge\sqrt{82}\)

Dấu "=" xảy ra <=> \(x=y=z=\frac{1}{3}\)