\(VT\le\frac{1}{\sqrt[3]{9}}\left(\frac{a+2b+3+3}{3}+\frac{b+2c+3+3}{3}+\frac{c+2a+3+3}{3}\right)\)
\(=\frac{1}{\sqrt[3]{9}}.\frac{3\left(a+b+c\right)+18}{3}=\frac{9}{\sqrt[3]{9}}=\sqrt[3]{81}=3\sqrt[3]{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)
Cho a,b,c>0 thỏa mãn a+b+c=3
\(\sqrt{2a+b}+\sqrt{2b+c}+\sqrt{2c+a}\le3\sqrt{3}\)
\(\sqrt[3]{a+b+c}+\sqrt[3]{b+c+1}+\sqrt[3]{c+a+1}\le3\sqrt[3]{3}\)
Cho a,b,c >0 và a+b+c = 3
cm \(\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\le3\sqrt[3]{3}\)
Cho 3 số thực dương a, b,c CMR \(\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}}\le3\)\(\le\)3
Cho \(a+b+c=3\).
CM
a)\(\sqrt[5]{2a+b}+\sqrt[5]{2b+c}+\sqrt[5]{2c+a}\le3\sqrt[5]{3}\)
b)\(\sqrt[5]{a\left(a+c\right)\left(2a+b\right)}+\sqrt[5]{b\left(b+a\right)\left(2b+c\right)}+\sqrt[5]{c\left(c+b\right)\left(2c+a\right)}\le3\sqrt[5]{6}\)
Cho các số dương a,b,c thoả mãn điều kiện a+b+c=3. Chứng minh rằng: \(\sqrt{2a^2+ab+2b^2}+\sqrt{2b^2+bc+2c^2}+\sqrt{2c^2+ca+2a^2}\ge3\sqrt{5}\)
cho a,b,c >0 và a+b+c=3
cmr \(\sqrt[5]{\left(2a+b\right)\left(a+c\right)a}+\sqrt[5]{\left(2b+c\right)\left(b+a\right)b}+\sqrt[5]{\left(2c+a\right)\left(c+b\right)c}\) \(\le3\sqrt[5]{6}\)
Cho a,b,c>0 CMR
\(a\sqrt{\frac{a}{a+2b}}+b\sqrt{\frac{b}{b+2c}}+c\sqrt{\frac{c}{c+2a}}\ge\frac{a+b+c}{\sqrt{3}}\)
1) cho a;b;c ko âm .chứng minh \(\sqrt{\frac{a+2b}{3}}+\sqrt{\frac{b+2c}{3}}+\sqrt{\frac{c+2a}{3}}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}\)
2) cho a;;b;c dương và abc=1. chứng minh \(\frac{bc}{a^2b+a^2c}+\frac{ca}{b^2c+b^2a}+\frac{ab}{c^2a+c^2b}\ge\frac{3}{2}\)
Cho a,b,c >0 và a+b+c=3
CMR: \(\frac{1}{\sqrt{2a+b+1}}+\frac{1}{\sqrt{2b+c+1}}+\frac{1}{\sqrt{2c+a+1}}\ge\frac{3}{2}\)
