Question

Cho x+y+z=10. CMR: \(x^2+y^2+z^2\ge\dfrac{100}{3}\)

Answer 1

Ta có:

\(\left(x-y\right)^2\ge0\)

\(\Rightarrow x^2+y^2\ge2xy\)

Tương tự có:

\(x^2+z^2\ge2xz\)

\(y^2+z^2\ge2yz\)

Do đó :

\(x^2+y^2+x^2+x^2+y^2+z^2\ge2xy+2xz+2yz\)

\(\Rightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\)

\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)+x^2+y^2+z^2\)

\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)

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

\(\Rightarrow x^2+y^2+z^2\ge\dfrac{10^2}{3}\) (thay x+y+z=10)

\(\Rightarrow x^2+y^2+z^2\ge\dfrac{100}{3}\)

\(\rightarrowđpcm\)