Chứng minh a^3/căn((a^3+1)(b^3+1))+b^2/căn((b^3+1)+..>=4/3

Áp dụng BDT AM-GM:

\(\sqrt{a^3+1}=\sqrt{\left(a+1\right)\left(a^2-a+1\right)}\le\dfrac{1}{2}\left(a^2+2\right)\)

tương tự như thế, ta có:

\(VT=\sum\dfrac{a^2}{\sqrt{\left(a^3+1\right)\left(b^3+1\right)}}\ge\sum\dfrac{4a^2}{\left(a^2+2\right)\left(b^2+2\right)}\)

Cần chứng minh \(\sum\dfrac{4a^2}{\left(a^2+2\right)\left(b^2+2\right)}\ge\dfrac{4}{3}\Leftrightarrow\sum\dfrac{a^2}{\left(a^2+2\right)\left(b^2+2\right)}\ge\dfrac{1}{3}\)

\(\Leftrightarrow3\left[a^2\left(c^2+2\right)+b^2\left(a^2+2\right)+c^2\left(b^2+2\right)\right]\ge\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)\)

\(\Leftrightarrow3\left[a^2b^2+b^2c^2+c^2a^2+2\left(a^2+b^2+c^2\right)\right]\ge a^2b^2c^2+2\left(a^2b^2+b^2c^2+c^2a^2\right)+4\left(a^2+b^2+c^2\right)+8\)

\(\Leftrightarrow\left(a^2b^2+b^2c^2+c^2a^2\right)+2\left(a^2+b^2+c^2\right)\ge72\)

Điều này đúng theo AM-GM:

\(\left\{{}\begin{matrix}a^2b^2+b^2c^2+c^2a^2\ge3\sqrt[3]{a^4b^4c^4}=48\\a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}=12\end{matrix}\right.\)(abc=8)

Vậy ta có đpcm.Dấu = xảy ra khi a=b=c=2

P/s : góp thêm đề :

1) \(\sum\sqrt{\dfrac{a}{b+3c}}\ge\dfrac{3}{2}\)

2)\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{2\left(ab+bc+ca\right)}{a^2+b^2+c^2}\ge5\)

3) \(a^3+b^3+c^3=3.\)CMR:\(\sum\dfrac{a}{b+2}\le1\)

4) CMr :\(\sum\dfrac{a^2}{a^2+b^2}\ge\sum\dfrac{a}{a+b}\)