Question

Với a,b \(\ge\) 0 chứng minh :

a, \(a^4+b^4+18\ge12ab\)

b, \(2\sqrt{\dfrac{a}{b}}+3\sqrt[3]{\dfrac{b}{a}}\ge5\)

Answer 1

a. - Áp dụng bất đẳng thức Caushy ta có:

\(a^4+b^4+9+9\ge4\sqrt[4]{a^4b^4.9.9}=4ab.3=12ab\)

\(\Rightarrow a^4+b^4+18\ge12ab\)

Dấu "=" xảy ra \(\Leftrightarrow a^4=b^4=9;a,b\ge0\Leftrightarrow a=b=\sqrt{3}\)

b. - Áp dụng BĐT Caushy ta có:

\(\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{a}{b}}+\sqrt[3]{\dfrac{b}{a}}+\sqrt[3]{\dfrac{b}{a}}+\sqrt[3]{\dfrac{b}{a}}\ge5\sqrt[5]{\left(\sqrt{\dfrac{a}{b}}\right)^2.\left(\sqrt[3]{\dfrac{b}{a}}\right)^3}=5\sqrt[5]{\dfrac{a}{b}.\dfrac{b}{a}}=5\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{\dfrac{a}{b}}=\sqrt[3]{\dfrac{b}{a}}\Leftrightarrow a=b>0\)