BĐT <=> \(3\left(1-\sqrt[3]{x}+\sqrt[3]{x^2}\right)\ge1+\sqrt[3]{x}+\sqrt[3]{x^2}\)
<=> \(3-3\sqrt[3]{x}+3\sqrt[3]{x^2}\ge1+\sqrt[3]{x}+\sqrt[3]{x^2}\)
<=> \(3-3\sqrt[3]{x}+3\sqrt[3]{x^2}-1-\sqrt[3]{x}-\sqrt[3]{x^2}\ge0\)
<=> \(2\sqrt[3]{x^2}-4\sqrt[3]{x}+2\ge0\)
<=> \(2\left(\sqrt[3]{x^2}-1\right)^2\ge0\) luôn đúng với mọi x => đpcm
\(\sqrt{X}+\sqrt{Y}+\sqrt{Z}=3\) 3 cmr \(X^2\sqrt{X}+y^2+\sqrt{Y}+z^2\sqrt{Z}+\frac{1}{\sqrt{X}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{Z}}\ge\)
Cho x,y,z>0,x+y+z=1.CMR
\(\frac{\sqrt{x}}{1-x}+\frac{\sqrt{y}}{1-y}+\frac{\sqrt{z}}{1-z}\ge\frac{3\sqrt{3}}{2}\)
Cho x,y,z là các số dương. Chứng minh rằng:
\(\frac{1}{\sqrt{x}+3\sqrt{y}}+\frac{1}{\sqrt{y}+3\sqrt{z}}+\frac{1}{\sqrt{z}+3\sqrt{x}}\ge\frac{1}{\sqrt{x}+2\sqrt{y}+\sqrt{z}}+\frac{1}{\sqrt{y}+2\sqrt{z}+\sqrt{x}}+\frac{1}{\sqrt{z}+2\sqrt{x}+\sqrt{y}}\)
Cho x, y, z là 3 số thực dương và x + y + z ≤ 1. CMR:
\(\sqrt{x^2+\frac{1}{y^2}}+\sqrt{y^2+\frac{1}{z^2}}+\sqrt{z^2+\frac{1}{x^2}}\ge\sqrt{82}\)
1) \(\frac{\sqrt{2\left(X^2-16\right)}}{\sqrt{X-3}}+\sqrt{X-3}>\frac{7-X}{\sqrt{X-3}}\)
2) \(\frac{1}{\sqrt{2X^2+3X-5}}\ge\frac{1}{2X-1}\)
3) \(\frac{1-\sqrt{1-4X^2}}{X}< 3\)
4) \(\frac{\sqrt{3X+1}-X}{2X-1}< 1\)
Cho 3 số dương x,y,z là 3 số dương và x+y+z=1. CMR: \(\sqrt{x^2+\frac{1}{x^2}}\)+\(\sqrt{y^2+\frac{1}{y^2}}\)+\(\sqrt{z^2+\frac{1}{z^2}}\)\(\ge\sqrt{82}\)
cho \(A=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\frac{3-\sqrt{x}}{\sqrt{x}+1}\right).\)
Tìm x để A \(\ge\frac{3}{2}\)
Giải phương trình:
\(a)\sqrt{x^2+2x+4}\ge x-2\\ b)x=\sqrt{x-\frac{1}{x}}+\sqrt{x+\frac{1}{x}}\\ c)\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2\sqrt{2x-5}}\\ d)x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\\ e)\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\)
Cho 3 số dương x,y,z. CMR:
\(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\)
>= \(3\left(\frac{1}{\sqrt{x}+2\sqrt{y}}+\frac{1}{\sqrt{y}+2\sqrt{z}}+\frac{1}{\sqrt{z}+2\sqrt{x}}\right)\)
