Question

Answer 1

k: \(\frac{a}{b}+\frac{b}{a}\ge2\cdot\sqrt{\frac{a}{b}\cdot\frac{b}{a}}=2\cdot\sqrt1=2\) ∀a,b>0

m: \(a^3+b^3\ge a^2b+ab^2\) (1)

=>\(a^3+b^3-a^2b-ab^2\ge0\)

=>\(a^2\left(a-b\right)-b^2\left(a-b\right)\ge0\)

=>\(\left(a-b\right)^2\cdot\left(a+b\right)\ge0\)

mà \(\left(a-b\right)^2\ge0;\left(a+b\right)>0\forall a,b>0\)

nên (1) đúng

s: \(x^2+y^2+z^2\ge xy+yz+xz\)

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

=>\(x^2-2xy+y^2+y^2-2yz+z^2+x^2-2xz+z^2\ge0\)

=>\(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\) (luôn đúng)