Theo AM - GM ta có:
\(x+y\ge2\sqrt{xy}\)
\(3y+3z\ge2\sqrt{3y.3z}=6\sqrt{yz}\)
\(5z+5x\ge2\sqrt{5z.5x}=10\sqrt{zx}\)
Cộng theo vế ta có: \(6x+4y+8z\ge2\sqrt{xy}+6\sqrt{yz}+10\sqrt{zx}\)
=> \(3x+2y+4z\ge\sqrt{xy}+3\sqrt{yz}+5\sqrt{zx}\)
Dấu "=" xảy ra <=> x = y = z
Cho a, b, c > 0 và x + y + z = 3 .
CMR : \(\dfrac{x}{x+\sqrt{3x+yz}}+\dfrac{y}{y+\sqrt{3y+zx}}+\dfrac{z}{z+\sqrt{3z+xy}}\le1\)
Cho x,y,z>0. CMR:
\(\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\ge\sqrt{3}\left(x+y+z\right)\)
Cho a,b,c>0, \(x^2+y^2+y^2=3\)
CMR: \(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xy}}+\frac{z}{\sqrt[3]{xy}}\ge xy+yz+zx\)
Cho x;y;z > 0 thỏa mãn x2 + y2 + z2 = 3
CMR: \(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\ge xy+yz+zx\)
CMR: \(\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\ge\sqrt{3}\left(x+y+z\right)\)
Với x;y;z > 0.
Cho x,y,z > 0 ; x + y + z = 1
CMR: \(\sqrt{\frac{xy}{z+xy}}+\sqrt{\frac{yz}{x+yz}}+\sqrt{\frac{zx}{y+zx}}\le\frac{3}{2}\)
cho x;y;z là các số dương thỏa mãn x+y+z=1.Chứng minh \(\sqrt{2x^2+xy+2y^2}+\sqrt{2y^2+yz+2z^2}+\sqrt{2z^2+zx+2x^2}\ge\sqrt{5}\)
Cho x,y,z >0 và x+y+z=3 CM: \(\sqrt{x}+\sqrt{y}+\sqrt{z}\ge xy+yz+zx\)
cho x,y,z>0 thoả mãn x2+y2+z2=3. Chứng minh rằng:
\(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\ge xy+yz+zx\)
