Với x,y,z dương
Ta có:(x-y)2\(\ge0\forall x;y\)
=>x2+y2\(\ge\)2xy
Dấu = xảy ra khi x=y
Tương tự y2+z2\(\ge\)2yz
z2+x2\(\ge\)2zx
Cộng vế với vế 3 BĐT =>2(x2+y2+z2)\(\ge\)2(xy+yz+zx)
<=>x2+y2+z2\(\ge\)xy+yz+zx
<=>\(\dfrac{3}{xy+yz+zx}\ge\dfrac{3}{x^2+y^2+z^2}\)
Dấu = xảy ra khi và chỉ khi x=y=z
=>\(\dfrac{3}{xy+yz+zx}+\dfrac{2}{x^2+y^2+z^2}\ge\dfrac{5}{x^2+y^2+z^2}\)
Áp dụng BĐT bunhiacopski:
\(\left(x^2+y^2+z^2\right)\left(\dfrac{1}{3^2}+\dfrac{1}{3^2}+\dfrac{1}{3^3}\right)\le\left(\dfrac{x+y+z}{3}\right)^2=\dfrac{1}{3^2}=\dfrac{1}{9}\)(Do x+y+z=1)
Dấu = xảy ra khi và chỉ khi \(\dfrac{x}{3}=\dfrac{y}{3}=\dfrac{z}{3}\)<=>x=y=z
=>\(\dfrac{5}{x^2+y^2+z^2}=\dfrac{5}{3\cdot\left(x^2+y^2+z^2\right)\left(\dfrac{1}{3^2}+\dfrac{1}{3^2}+\dfrac{1}{3^2}\right)}\ge\dfrac{5}{3\cdot\dfrac{1}{9}}=15\)
=>\(\dfrac{3}{xy+yz+zx}+\dfrac{2}{x^2+y^2+z^2}\ge15\)(đpcm)
Dấu = xảy ra khi \(\left\{{}\begin{matrix}x=y=z\\z+y+z=1\end{matrix}\right.\)<=>x=y=z=\(\dfrac{1}{3}\)
