Ta có :
\(a-\sqrt{a}+\frac{1}{4}=\left(\sqrt{a}-\frac{1}{2}\right)^2\ge0\forall a\ge0\Rightarrow a+\frac{1}{4}\ge\sqrt{a}\)
\(b-\sqrt{b}+\frac{1}{4}=\left(\sqrt{b}-\frac{1}{2}\right)^2\ge0\forall b\ge0\Rightarrow b+\frac{1}{4}\ge\sqrt{b}\)
\(\Rightarrow a+\frac{1}{4}+b+\frac{1}{4}\ge\sqrt{a}+\sqrt{b}\)
\(\Rightarrow a+b+\frac{1}{2}\ge\sqrt{a}+\sqrt{b}\)(đpcm)
vs \(a\ge0;b\ge0\)
cm \(\sqrt{\frac{a+b}{2}}\ge\frac{\sqrt{a}+\sqrt{b}}{2}\)
Cho \(a\ge0\), \(b\ge0\). CMR: \(\frac{1}{2}\left(a+b\right)^2+\frac{1}{4}\left(a+b\right)\ge a\sqrt{b}+b\sqrt{a}\)
\(CMR:\sqrt[3]{\frac{a^3+b^3}{2}}\ge\sqrt{\frac{a^2+b^2}{2}};a\ge0;b\ge0\)
Mình đi học thội!
PP............
Rút gọn
a)\(2\sqrt{a}+3a\sqrt{4ab^2}-2b\sqrt{16a^5}-2\sqrt{25a}\)(a>0;b>0)
b)\(\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\left(a\ge0;b\ge0;a\ne b\right)\)
c)\(\frac{a\sqrt{a}-b\sqrt{b}}{a-b}-\frac{a-b}{\sqrt{a}-\sqrt{b}}\left(a\ge0;b\ge0;a\ne0\right)\)
\(cho\)\(a\ge0\),\(b\ge0\)
\(cm\) \(\frac{a+b}{2}\) \(\ge\sqrt{ab}\)
b1 sử dụng HDT hoặc co-si
a)cho x\(\ge\)0,y\(\ge\)1,z\(\ge\)2cmr \(x\sqrt{y-1}+y\sqrt{x-1}\le xy\)
b)cho \(x\ge0,y\ge1,z\ge2cmr\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}\le\frac{1}{2}\left(x+y+z\right)\)
c)cho a,b,c\(\ge0\)cmr \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\)
Cho \(a,b\ge0\). Chứng minh \(\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}\ge\sqrt{a}+\sqrt{b}\)
baaif 1 chứng tỏ
\(\sqrt{a+\sqrt{b}}\ge\sqrt{a+b}\) với \(a\ge0,b\ge0\)
Cmr \(\sqrt{a^2+b^2}\ge\frac{a+b}{\sqrt{2}}\text{với mọi}a;b\ge0\)
