Question

[๖ۣۜLuyri Vũ๖ۣۜ]

Giúp em với ạ.

Answer 1

[Đoàn Đức Hà]

Ta có: \(\sqrt{1+x^2+y^2}\ge\sqrt{3\sqrt[3]{x^2y^2}}=\sqrt{3}\sqrt[3]{xy}\)

\(\Rightarrow\frac{\sqrt{1+x^2+y^2}}{xy}\ge\frac{\sqrt{3}\sqrt[3]{xy}}{xy}=\sqrt{3}\sqrt[3]{xy}z\)

Tương tự ta cũng có: \(\frac{\sqrt{1+y^2+z^2}}{yz}\ge\sqrt{3}\sqrt[3]{yz}x,\frac{\sqrt{1+z^2+x^2}}{zx}\ge\sqrt{3}\sqrt[3]{zx}y\)

Suy ra \(P\ge\sqrt{3}\left(\sqrt[3]{xy}z+\sqrt[3]{yz}x+\sqrt[3]{zx}y\right)\ge3\sqrt{3}\sqrt[3]{\sqrt[3]{x^5y^5z^5}}=3\sqrt{3}\)

Dấu \(=\)khi \(x=y=z=1\).

Answer 2

[Phan Nghĩa]

Áp dụng BĐT Svacxo ta có :

\(P\ge sigma\frac{\sqrt{\left(1+x+y\right)^2}}{\sqrt{3}.xy}=\frac{z+xz+yz}{\sqrt{3}xyz}+\frac{x+xy+xz}{\sqrt{3}xyz}+\frac{y+yz+yx}{\sqrt{3}xyz}\)

\(=\frac{x+y+z+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}{\sqrt{3}}\)

Theo AM-GM thì \(\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge6+3\sqrt[3]{\frac{1}{xyz}}=9\)

Khi đó : \(\frac{x+y+z+\frac{2}{x}+\frac{2}{y}+\frac{2}{z}}{\sqrt{3}}\ge\frac{9}{\sqrt{3}}=3\sqrt{3}\)

Dấu "=" xảy ra khi và chỉ khi \(x=y=z=1\)

Vậy Min P = 3 căn 3 khi x=y=z=1 

okela -))