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Question

Cho x, y, z > 0 và x + y + z = 1. Chứng minh rằng: $\sqrt{2x^2+xy+2y^2}+\sqrt{2y^2+yz+2z^2}+\sqrt{2z^2+zx+2x^2}\ge\sqrt{5}$

Nguyễn Việt Lâm · Accepted Answer

$VT=\sum\sqrt{\frac{1}{2}\left(x^2+2xy+y^2\right)+\frac{3}{2}\left(x^2+y^2\right)}$

$VT\ge\sum\sqrt{\frac{1}{2}\left(x+y\right)^2+\frac{3}{4}\left(x+y\right)^2}=\sum\sqrt{\frac{5}{4}\left(x+y\right)^2}$

$VT\ge\frac{\sqrt{5}}{2}\left(x+y\right)+\frac{\sqrt{5}}{2}\left(y+z\right)+\frac{\sqrt{5}}{2}\left(z+x\right)$

$VT\ge\sqrt{5}\left(x+y+z\right)=\sqrt{5}$

Dấu "=" xảy ra khi $x=y=z=\frac{1}{3}$Câu trả lời: $VT=\sum\sqrt{\frac{1}{2}\left(x^2+2xy+y^2\right)+\frac{3}{2}\left(x^2+y^2\right)}$

$VT\ge\sum\sqrt{\frac{1}{2}\left(x+y\right)^2+\frac{3}{4}\left(x+y\right)^2}=\sum\sqrt{\frac{5}{4}\left(x+y\right)^2}$

$VT\ge\frac{\sqrt{5}}{2}\left(x+y\right)+\frac{\sqrt{5}}{2}\left(y+z\right)+\frac{\sqrt{5}}{2}\left(z+x\right)$

$VT\ge\sqrt{5}\left(x+y+z\right)=\sqrt{5}$

Dấu "=" xảy ra khi $x=y=z=\frac{1}{3}$