\(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)

\(B\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)

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

\(B\ge\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)

\(B\ge\sqrt{\left(x+y+z\right)^2+\dfrac{162}{\left(x+y+z\right)^2}}\)

\(B\ge\sqrt{4+\dfrac{162}{4}}=\sqrt{\dfrac{89}{2}}\)

Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)

cách làm 

700 : 100*7 =49

 đáp số:49

Áp dụng liên tiếp bất đẳng thức CauChy-Schwarz và AM-GM\(A=\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\)

\(A=1+\dfrac{1}{b}+\dfrac{1}{a}+\dfrac{1}{ab}\)

\(A\ge1+\dfrac{\left(1+1\right)^2}{a+b}+\dfrac{1}{\dfrac{\left(a+b\right)^2}{4}}\)

\(A\ge1+\dfrac{4}{a+b}+\dfrac{4}{\left(a+b\right)^2}=1+4+4=9\)

Dấu "=" xảy ra khi: \(a=b=\dfrac{1}{2}\)