Áp dụng BĐT AM-GM ta có:
\(\frac{a}{\sqrt{b}}+\sqrt{b}\ge2.\sqrt{\frac{a}{\sqrt{b}}.\sqrt{b}}=2\sqrt{a}\)
Tương tự:\(\frac{b}{\sqrt{a}}+\sqrt{a}\ge2\sqrt{\frac{b}{\sqrt{a}}.\sqrt{a}}=2\sqrt{b}\)
Cộng theo vế BĐT ta được:\(\frac{a}{\sqrt{b}}+\sqrt{b}+\frac{b}{\sqrt{a}}+\sqrt{a}\ge2\left(\sqrt{a}+\sqrt{b}\right)\)
\(\Rightarrow\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}\ge\sqrt{a}+\sqrt{b}\)
Cho a>0, b>0. CMR:
\(\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}\ge\sqrt{a}+\sqrt{b}\)
Cho a,b,c>0 và abc=1
cmr: \(\frac{b+c}{\sqrt{a}}+\frac{a+c}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3
\)
Với \(a,b>0.CMR:\frac{a}{\sqrt{b}}-\sqrt{a}\ge\sqrt{b}-\frac{b}{\sqrt{a}}\)
Cho a,b,c>0 CMR
\(a\sqrt{\frac{a}{a+2b}}+b\sqrt{\frac{b}{b+2c}}+c\sqrt{\frac{c}{c+2a}}\ge\frac{a+b+c}{\sqrt{3}}\)
Cho a,b,c >0 và abc= 1.CMR:
\(\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)
Giúp với , cần gấp
cho a,b,c >0
cmr \(\sqrt{\frac{a+b}{c+ab}}+\sqrt{\frac{b+c}{a+bc}}+\sqrt{\frac{c+a}{b+ca}}\ge\)3
Cho a,b,c>0. Cmr:
\(\frac{a}{\sqrt{ab+b^2}}+\frac{b}{\sqrt{bc+b^2}}+\frac{c}{\sqrt{ac+c^2}}\ge\frac{3\sqrt{2}}{2}\)
CMR: với a, b, c > 0 thì:
\(\sqrt{\frac{a}{bc}}+\sqrt{\frac{b}{ca}}+\sqrt{\frac{c}{ab}}\ge\sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}}\)
Cho a b c > 0 và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\). CMR \(\sqrt[4]{a^3}+\sqrt[4]{b^3}+\sqrt[4]{c^3}\ge\sqrt[3]{a^2}+\sqrt[3]{b^2}+\sqrt[3]{c^2}\)
