Do vai trò của 3 biến là như nhau, ko mất tính tổng quát, giả sử \(a\ge b\ge c\)

\(\Rightarrow\) Theo BĐT Chebyshev:

\(3\left(a^3+b^3+c^3\right)\ge\left(a^2+b^2+c^2\right)\left(a+b+c\right)\) (1)

Bunhiacopxki:

\(\left(a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\right)^2\le2\left(a^2+b^2+c^2\right)\left(a+b+c\right)\le6\left(a^3+b^3+c^3\right)\)

Nên ta chỉ cần chứng minh:

\(\left(a^3+b^3+c^3\right)^2\ge6\left(a^3+b^3+c^3\right)\)

\(\Leftrightarrow a^3+b^3+c^3\ge6\)

Hiển nhiên đúng do: \(a^3+b^3+c^3\ge3abc=6\)

Ta thấy: \(\Sigma_{cyc}\sqrt[3]{\frac{a^2+bc}{abc\left(b^2+c^2\right)}}=\Sigma_{cyc}\frac{a^2+bc}{\sqrt[3]{\left(a^2b+b^2c\right)\left(bc^2+ca^2\right)\left(c^2a+ab^2\right)}}\)

Ta lại có: \(\sqrt[3]{\left(a^2b+b^2c\right)\left(bc^2+ca^2\right)\left(c^2a+ab^2\right)}\le\frac{\left(a^2b+b^2c\right)+\left(bc^2+ca^2\right)+\left(c^2a+ab^2\right)}{3}=\frac{1}{3}\Sigma_{cyc}\left(ab\left(a+b\right)\right)\)

\(\Leftrightarrow\Sigma_{cyc}\sqrt[3]{\frac{a^2+bc}{abc\left(b^2+c^2\right)}}\ge\frac{\Sigma_{cyc}\left(a^2+bc\right)}{\frac{1}{3}\Sigma_{cyc}\left(ab\left(a+b\right)\right)}=\frac{a^2+b^2+c^2+ab+bc+ca}{\frac{1}{3}\Sigma_{cyc}\left(ab\left(a+b\right)\right)}\)

Nhận thấy: \(A=\left(a+b+c\right)\left(a^2+b^2+c^2+ab+bc+ca\right)=a^3+b^3+c^3+3abc+2\Sigma_{cyc}\left(ab\left(a+b\right)\right)\)

Theo Schur: \(a^3+b^3+c^3+3abc\ge\Sigma_{cyc}\left(ab\left(a+b\right)\right)\)

\(\Leftrightarrow A\ge3\Sigma_{cyc}\left(ab\left(a+b\right)\right)\)

\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\frac{a^2+bc}{abc\left(b^2+c^2\right)}}\ge\frac{3\Sigma_{cyc}\left(ab\left(a+b\right)\right)}{\frac{1}{3}\left(a+b+c\right)\Sigma_{cyc}\left(ab\left(a+b\right)\right)}=\frac{9}{a+b+c}\)

\(VP^2\le2\left(a+b+c\right)\left(a^2+b^2+c^2\right)\) (1) 

\(VT^2=\left(\frac{a^4}{a}+\frac{b^4}{b}+\frac{c^4}{c}\right)^2\ge\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)^3}{\left(a+b+c\right)^2}\)

\(\ge\frac{\left(a^2+b^2+c^2\right)\left(a+b+c\right)^6}{27\left(a+b+c\right)^2}=\frac{\left(a^2+b^2+c^2\right)\left(a+b+c\right)\left(a+b+c\right)^3}{27}\)

\(\ge\frac{\left(a+b+c\right)\left(a^2+b^2+c^2\right)\left(3\sqrt[3]{abc}\right)^3}{27}=2\left(a+b+c\right)\left(a^2+b^2+c^2\right)\ge VP^2\) (2) 

Mà VT và VP đều dường nên từ (1) và (2) suy ra đpcm 

Dấu "=" xảy ra khi \(a=b=c=\sqrt[3]{2}\)

\(\dfrac{a^3}{1+b}+\dfrac{1+b}{4}+\dfrac{1}{2}\ge3\sqrt[3]{\dfrac{a^3\left(1+b\right)}{8\left(a+b\right)}}=\dfrac{3a}{2}\)

\(\dfrac{b^3}{1+c}+\dfrac{1+c}{4}+\dfrac{1}{2}\ge\dfrac{3b}{2}\) ; \(\dfrac{c^3}{1+a}+\dfrac{1+a}{4}+\dfrac{1}{2}\ge\dfrac{3c}{2}\)

\(\Rightarrow VT+\dfrac{a+b+c}{4}+\dfrac{9}{4}\ge\dfrac{3}{2}\left(a+b+c\right)\)

\(\Rightarrow VT\ge\dfrac{5}{4}\left(a+b+c\right)-\dfrac{9}{4}\ge\dfrac{5}{4}.3\sqrt[3]{abc}-\dfrac{9}{4}=\dfrac{3}{2}\)