Sử dụng BĐT: \(\left(x+y+z\right)^3\ge27xyz\Rightarrow\left(\frac{x+y+z}{3}\right)^3\ge xyz\)

\(\Rightarrow\left(\frac{1+a+1+b+1+c}{3}\right)^3\ge\left(1+a\right)\left(1+b\right)\left(1+c\right)\)

Ta có: \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge3\sqrt[3]{\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

\(\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}\ge3\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

Cộng vế với vế:

\(1\ge\frac{1+\sqrt[3]{abc}}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\Rightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)

Dấu "=" 3 BĐT trên xảy ra khi \(a=b=c\)

Lại có:

\(1+\sqrt[3]{abc}\ge2\sqrt{\sqrt[3]{abc}}\Rightarrow\left(1+\sqrt[3]{abc}\right)^3\ge\left(2\sqrt{\sqrt[3]{abc}}\right)^3=8\sqrt{abc}\)Dấu "=" xảy ra khi \(a=b=c=1\)

Bất đẳng thức cần chứng minh tương đương:

\(a^{10}b^2+b^{10}a^2\ge a^8b^4+b^8a^4\)

\(\Leftrightarrow a^8+b^8\ge a^6b^2+b^6a^2\) (Do \(a^2b^2\ge0\))

\(\Leftrightarrow\left(a^6-b^6\right)\left(a^2-b^2\right)\ge0\)

\(\Leftrightarrow\left(a^2-b^2\right)^2\left(a^4+a^2b^2+b^4\right)\ge0\) (luôn đúng).

Vậy ta có đpcm.

\(a^8+b^8-a^6b^2-a^2b^6=\left(a^8-a^6b^2\right)+\left(b^8-a^2b^6\right)=a^6\left(a^2-b^2\right)+b^6\left(b^2-a^2\right)=\left(a^6-b^6\right)\left(a^2-b^2\right)\) nên suy ra được như vậy Quỳnh Anh

Nhận xét thấy : \(x^4+y^4+z^4+t^4\ge2x^2y^2+2z^2t^2\ge4xyzt\)

Dấu " =" xảy ra khi \(x=y=z=t\)

Áp dụng :

\(a^4+a^4+b^4+c^4\ge4a^2bc\)

\(a^4+b^4+b^4+c^4\ge4ab^2c\)

\(a^4+b^4+c^4+c^4\ge4abc^2\)

\(\Rightarrow4\left(a^4+b^4+c^4\right)\ge4abc\left(a+b+c\right)\)

\(\Leftrightarrowđpcm\)

Dấu "  = " xảy ra khi \(a=b=c\)

a ) CM : \(a^4+b^4\ge a^3b+b^3a\)

Giả sử điều cần c/m là đúng

\(\Rightarrow a^4+b^4-a^3b-b^3a\ge0\)

\(\Rightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)

\(\Rightarrow\left(a^3-b^3\right)\left(a-b\right)\ge0\)

\(\Rightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)

Ta có : \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\a^2+ab+b^2=\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}\ge0\end{matrix}\right.\)

\(\Rightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)

\(\Rightarrow a^4+b^4-a^3b-b^3a\ge0\)

\(\Rightarrow a^4+b^4\ge a^3b+b^3a\)

\(\Rightarrow2\left(a^4+b^4\right)\ge a^4+a^3b+b^4+b^3a\)

\(\Rightarrow2\left(a^4+b^4\right)\ge\left(a+b\right)\left(a^3+b^3\right)\)

\(\left(đpcm\right)\)

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

\(=a^4+a^3b+a^3c+b^3a+b^4+b^3c+c^3a+c^3b+c^4\)

\(=\left(a^4+b^4+c^4\right)+\left(a^3b+b^3a\right)+\left(b^3c+c^3b\right)+\left(a^3c+c^3a\right)\)

CMTT như a ) : \(\left\{{}\begin{matrix}a^4+b^4\ge a^3b+b^3a\\b^4+c^4\ge b^3c+c^3b\\a^4+c^4\ge a^3c+c^3a\end{matrix}\right.\)

\(\Rightarrow2\left(a^4+b^4+c^4\right)\ge a^3b+b^3a+b^3c+c^3b+a^3c+c^3a\)

\(\Rightarrow3\left(a^4+b^4+c^4\right)\ge a^4+b^4+c^4+a^3b+b^3a+b^3c+c^3b+a^3c+c^3a\)

\(\Rightarrow3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\left(đpcm\right)\)

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