Vì x,y,z>0 nên áp dung bất đẳng thức Cô-si ta có:

\(\dfrac{1}{x^2+2yz}\)+\(\dfrac{1}{y^2+2xz}\)+\(\dfrac{1}{z^2+2xy}\)≥\(\dfrac{\left(1+1+1\right)^3}{x^2+y^2+z^2+2xy+2yz+2xz}\)

mà x+y+z=1 ⇔ x²+y²+z²+2xy+2yz+2zx=1 (bình phương cả 2 vế)

nên \(\dfrac{1}{x^2+2yz}\)+\(\dfrac{1}{y^2+2xz}\)+\(\dfrac{1}{z^2+2xy}\)≥\(\dfrac{\left(1+1+1\right)^2}{\left(x+y+z\right)^2}\)=9