Theo BĐT cô-si ta có:
\(\dfrac{a+b}{2}\ge\sqrt{ab}\Rightarrow ab\le\left(\dfrac{a+b}{2}\right)^2=\left(\dfrac{4}{2}\right)^2=4\)
\(a^3+b^3\ge2\sqrt{a^3b^3}=2\sqrt{4^3}=16\)
\(a^4+b^4\ge2\sqrt{a^4b^4}=2\sqrt{4^4}=32\)
Ta có:
\(\dfrac{a}{8+2b^3}+\dfrac{b}{8+2a^3}\)
\(=\dfrac{a\left(8+2a^3\right)}{\left(8+2b^3\right)\left(8+2a^3\right)}+\dfrac{b\left(8+2b^3\right)}{\left(8+2b^3\right)\left(8+2a^3\right)}\)
\(=\dfrac{8a+2a^4+8b+2b^4}{\left(8+2b^3\right)\left(8+2a^3\right)}\)
\(=\dfrac{8\left(a+b\right)+2\left(a^4+b^4\right)}{64+16a^3+16b^3+4a^3b^3}\)
\(=\dfrac{8\left(a+b\right)+2\left(a^4+b^4\right)}{64+16\left(a^3+b^3\right)+4\cdot\left(ab\right)^3}\ge\dfrac{8\cdot4+2\cdot32}{64+16\cdot16+4\cdot4^3}=\dfrac{96}{576}=\dfrac{1}{6}\left(dpcm\right)\)
Ta có:
\(\dfrac{a}{8+2b^3}=\dfrac{4-b}{8+2b^3}\)
Ta sẽ chứng minh \(\dfrac{4-b}{8+2b^3}\ge-\dfrac{1}{8}b+\dfrac{1}{3}\left(\text{*}\right)\) với mọi \(b>0\)
Thật vậy, BĐT \(\left(\text{*}\right)\) tương đương với:
\(\dfrac{4-b}{4+b^3}+\dfrac{b-2}{4}\ge\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{b^4-2b^3+8}{4\left(4+b^3\right)}\ge\dfrac{1}{6}\Leftrightarrow b^4-2b^3+8\ge\dfrac{2}{3}\left(4+b^3\right)\)
\(\Leftrightarrow3b^4-8b^3+16\ge0\)
\(\Leftrightarrow\left(b-2\right)^2\left(3b^2+4b+4\right)\ge0\), đúng với mọi \(b>0\)
Áp dụng BĐT \(\left(\text{*}\right)\), ta có:
\(\dfrac{a}{8+2b^3}+\dfrac{b}{8+2a^3}\ge\dfrac{2}{3}-\dfrac{1}{8}\left(a+b\right)=\dfrac{1}{6}\), đpcm
Đẳng thức xảy ra khi và chỉ khi \(a=b=2\)
\(a,b>0;a+b=4\)
\(\sum\dfrac{a}{8+2b^3}=\dfrac{1}{8}\sum\dfrac{8a}{8+2b^3}=\dfrac{1}{8}\sum\left(a-\dfrac{2ab^3}{8+b^3+b^3}\right)\ge\dfrac{1}{8}\sum\left(a-\dfrac{2ab^3}{6b^2}\right)\)
\(=\dfrac{1}{8}\left(a+b\right)-\dfrac{1}{12}ab=\dfrac{1}{8}.4-\dfrac{1}{12}ab\ge\dfrac{1}{2}-\dfrac{1}{12}.\dfrac{\left(a+b\right)^2}{4}=\dfrac{1}{2}-\dfrac{1}{12}.\dfrac{4^2}{4}=\dfrac{1}{6}\left(dpcm\right)\)
Dấu "=" xảy ra khi \(a=b=2\)
