Cho a,b,c >0 và a+b+c = 3
cm \(\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\le3\sqrt[3]{3}\)
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5 tháng 7 2019
\(VT\le\frac{1}{\sqrt[3]{9}}\left(\frac{a+2b+3+3}{3}+\frac{b+2c+3+3}{3}+\frac{c+2a+3+3}{3}\right)\)
\(=\frac{1}{\sqrt[3]{9}}.\frac{3\left(a+b+c\right)+18}{3}=\frac{9}{\sqrt[3]{9}}=\sqrt[3]{81}=3\sqrt[3]{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)
DT
5 tháng 1 2018
Áp dụng BĐT phụ:
\(3\left(a^2+a^2+b^2\right)\ge\left(2a+b\right)^2\)
P=\(\sum\dfrac{a}{\sqrt{2a^2+b^2}+\sqrt{3}}\)
\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P=\sum\dfrac{a}{\sqrt{3\left(a^2+a^2+b^2\right)}+3}\)
\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P\le\sum\dfrac{a}{\sqrt{\left(2a+b\right)^2}+a+b+c}=\sum\dfrac{a}{3a+2b+c}\)
Xét M=\(\sum\dfrac{a}{3a+2b+c}\)
\(3-3M=\sum\dfrac{2b+c}{3a+2b+c}\)
\(\Rightarrow\)\(3-3M=\sum\dfrac{\left(2b+c\right)^2}{\left(3a+2b+c\right)\left(2b+c\right)}\ge\)\(\dfrac{\left(3a+3b+3c\right)^2}{\sum\left(3a+2b+c\right)\left(2b+c\right)}\)
Mà
\(\sum\left(3a+2b+c\right)\left(2b+c\right)=5a^2+5b^2+5c^2+13ab+13bc+13ac=5\left(a+b+c\right)^2+3\left(ab+bc+ac\right)\le5\left(a+b+c\right)^2+\left(a+b+c\right)^2\)
\(\Rightarrow\)\(3-3M\ge\dfrac{\left(3a+3b+3c\right)^2}{6\left(a+b+c\right)^2}\ge\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrow\)\(M\le\dfrac{1}{2}\)
\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P\le\dfrac{1}{2}\Rightarrow P\le\dfrac{\sqrt{3}}{2}\)
5 tháng 10 2017
a/ \(\sqrt[5]{2a+b}+\sqrt[5]{2b+c}+\sqrt[5]{2c+a}\)
\(=\frac{1}{\sqrt[5]{3^4}}\left(\sqrt[5]{3^4}.\sqrt[5]{2a+b}+\sqrt[5]{3^4}.\sqrt[5]{2b+c}+\sqrt[5]{3^4}.\sqrt[5]{2c+a}\right)\)
\(\le\frac{1}{\sqrt[5]{3^4}}\left(\frac{3+3+3+3+2a+b}{5}+\frac{3+3+3+3+2b+c}{5}+\frac{3+3+3+3+2c+a}{5}\right)\)
\(=\frac{1}{\sqrt[5]{3^4}}\left(\frac{36}{5}+\frac{3\left(a+b+c\right)}{5}\right)\)
\(=\frac{1}{\sqrt[5]{3^4}}.9=3\sqrt[5]{3}\)
G
23 tháng 5 2019
\(A=\frac{a\sqrt{a}}{\sqrt{a+b+2c}}+\frac{b\sqrt{b}}{\sqrt{b+c+2a}}+\frac{c\sqrt{c}}{\sqrt{c+a+2b}}\)
\(A=\frac{a^2}{\sqrt{a\left(a+b+2c\right)}}+\frac{b^2}{\sqrt{b\left(b+c+2a\right)}}+\frac{c^2}{\sqrt{c\left(c+a+2b\right)}}\)
\(\ge\frac{\left(a+b+c\right)^2}{\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}}\)
Xét: \(2\left(\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}\right)\)
\(=\sqrt{4a\left(a+b+2c\right)}+\sqrt{4b\left(b+c+2a\right)}+\sqrt{4c\left(c+a+2b\right)}\)
\(\le\frac{4a+a+b+2c+4b+b+c+2a+4c+c+a+2b}{2}=4\left(a+b+c\right)\)
\(\Rightarrow\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}\le2\left(a+b+c\right)\)
\(\Rightarrow\frac{\left(a+b+c\right)^2}{\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{3}{2}\)
\("="\Leftrightarrow a=b=c=1\)
TN
30 tháng 6 2017
Đặt \(x=\sqrt{\frac{b}{a}};y=\sqrt{\frac{c}{b}};z=\sqrt{\frac{a}{c}}\) thì \(xyz=1\) và BĐT cần chứng minh là
\(\sqrt{\frac{2}{x^2+1}}+\sqrt{\frac{2}{y^2+1}}+\sqrt{\frac{2}{z^2+1}}\le3\)
Giả sử \(x\le y\le z\Rightarrow\hept{\begin{cases}xy\le1\\z\ge1\end{cases}}\) ta có:
\(\left(\sqrt{\frac{2}{x^2+1}}+\sqrt{\frac{2}{y^2+1}}\right)^2\le2\left(\frac{2}{x^2+1}+\frac{2}{y^2+1}\right)\)
\(=4\left[1+\frac{1-x^2y^2}{\left(x^2+1\right)\left(y^2+1\right)}\right]\)
\(\le4\left[1+\frac{1-x^2y^2}{\left(xy+1\right)^2}\right]=\frac{8}{xy+1}=\frac{8z}{z+1}\)
\(\Rightarrow\sqrt{\frac{2}{x^2+1}}+\sqrt{\frac{2}{y^2+1}}\le2\sqrt{\frac{2z}{z+1}}\)
Nên còn phải chứng minh \(2\sqrt{\frac{2z}{z+1}}+\frac{2}{z+1}\le3\)
\(\Leftrightarrow1+3z-2\sqrt{2z\left(z+1\right)}\ge0\Leftrightarrow\left(\sqrt{2z}-\sqrt{z+1}\right)^2\ge0\)
BĐT cuối đúng hay ta có ĐPCM
Áp dụng BĐT Cô -si cho 3 số không âm là a+ 2b, 3,3, ta được:
\(\sqrt[3]{a+2b}=\frac{1}{\sqrt[3]{9}}\sqrt[3]{3.3\left(a+2b\right)}\le\frac{1}{\sqrt[3]{9}}.\frac{3+3+\left(a+2b\right)}{3}\)
\(=\frac{6+a+2b}{3\sqrt[3]{9}}\)
Tương tự ta có: \(\sqrt[3]{b+2c}\le\frac{6+b+2c}{3\sqrt[3]{9}}\); \(\sqrt[3]{c+2a}\le\frac{6+c+2a}{3\sqrt[3]{9}}\)
\(\Rightarrow\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\le\frac{18+3\left(a+b+c\right)}{3\sqrt[3]{9}}\)
\(=\frac{27}{3\sqrt[3]{9}}=3\sqrt[3]{3}\)
(Dấu "="\(\Leftrightarrow a=b=c=1\))