Ta có BĐT sau:

\(\frac{a^4+b^4}{a^3+b^3}\ge\frac{a+b}{2}\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\left(true\right)\)

Khi đó tương tự ta có nốt \(\frac{b^4+c^4}{b^3+c^3}\ge\frac{b+c}{2};\frac{c^4+a^4}{c^3+a^3}\ge\frac{c+a}{2}\)

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

Ta dễ chứng minh được 

\(\frac{a^4}{a^3+b^3}+\frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}=\frac{b^4}{a^3+b^3}+\frac{c^4}{b^3+c^3}+\frac{a^4}{a^3+c^3}\)( trừ cái là xong )

Khi đó \(LHS\ge\frac{a+b+c}{2}\)

Ta có điều phải chứng minh

Đẳng thức xảy ra tại a=b=c

Sử dụng BĐT Cauchu Schawrz cũng được

(a+b+c)(a³+b³+c³)

=a⁴+a³b+a³c+ab³+b⁴+b³c+ac³+bc³+c⁴

=a⁴+b⁴+c⁴+(a³b+ab³)+(bc³+b³c)+(c³a+ca³)

=a⁴+b⁴+c⁴+ab(a²+b²)+bc(b²+c²)+ca(c²+a²)

=(a⁴+b⁴+c⁴)+ab(a²+b²)+bc(b²+c²)+ca(c²+a²)

P/s đến đây bạn áp đụng bđt thức bunhi a là ra

(a+b+c) (a³+b³+c³)

=a⁴+a³b+a³c+ab³+b⁴+b³c+ac³+bc³+c⁴

=a⁴+b⁴+c⁴+(a³b+ab³)+(bc³+b³c)+(c³a+ca³)

=a⁴+b⁴+c⁴+ab(a²+b²)+bc(b²+c²)+ca(c²+a²)

=(a⁴+b⁴+c⁴)+ab(a²+b²)+bc(b²+c²)+ca(c²+a²)

\(\frac{a^4}{b+c}+\frac{b^4}{c+a}+\frac{c^4}{a+b}=\frac{a^6}{a^2b+a^2c}+\frac{b^6}{b^2a+b^2c}+\frac{c^6}{c^2a+c^2b}\ge\frac{\left(a^3+b^3+c^3\right)^2}{ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}\ge\frac{\left(a^3+b^3+c^3\right)^2}{2\left(a^3+b^3+c^3\right)}=\frac{a^3+b^3+c^3}{2}\)

Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)

\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)

CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)

Cộng vế theo vế:

\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)

Dấu \("="\Leftrightarrow a=b=c=2\)

Ta có:

\(\dfrac{3}{a}+\dfrac{3}{b}\ge\dfrac{12}{a+b}\) (1)

\(\Leftrightarrow\dfrac{3a\left(a+b\right)+3b\left(a+b\right)-12ab}{ab\left(a+b\right)}\ge0\)

\(\Leftrightarrow\dfrac{3a^2+3ab+3ab+3b^2-12ab}{ab\left(a+b\right)}\ge0\)

\(\Leftrightarrow\dfrac{3a^2+3b^2-6ab}{ab\left(a+b\right)}\ge0\)

\(\Leftrightarrow\dfrac{3\left(a-b\right)^2}{ab\left(a+b\right)}\ge0\) ( luôn đúng)

Tương tự ta có:

\(\dfrac{2}{b}+\dfrac{2}{c}\ge\dfrac{8}{b+c}\) (2)

\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\) (3)

Cộng vế (1) (2)(3) ta được:

\(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{2}{b}+\dfrac{2}{c}+\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{12}{a+b}+\dfrac{8}{b+c}+\dfrac{4}{c+a}\)

\(\Leftrightarrow\dfrac{4}{a}+\dfrac{5}{b}+\dfrac{3}{c}\ge4\left(\dfrac{3}{a+b}+\dfrac{2}{b+c}+\dfrac{1}{c+a}\right)\)

Lời giải:
Ta có:

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

\(=27-3(3-a)(3-b)(3-c)\)

\(=27-3[27-9(a+b+c)+3(ab+bc+ac)-abc]\)

\(=27-3[3(ab+bc+ac)-abc]=27-9(ab+bc+ac)+3abc\)

Do đó:

\(A=a^3+b^3+c^3+\frac{15}{4}abc=27-9(ab+bc+ac)+\frac{27}{4}abc(*)\)

Áp dụng BĐT Schur :

\(abc\geq (a+b-c)(b+c-a)(c+a-b)\)

\(\Leftrightarrow abc\geq (3-2a)(3-2b)(3-2c)\)

\(\Leftrightarrow abc\geq 27-18(a+b+c)+12(ab+bc+ac)-8abc\)

\(\Leftrightarrow 9abc\geq 12(ab+bc+ac)-27\)

\(\Leftrightarrow 3abc\geq 4(ab+bc+ac)-9\)

\(\Rightarrow \frac{27}{4}abc\geq 9(ab+bc+ac)-\frac{81}{4}(**)\)

Từ \((*); (**)\Rightarrow A\geq 27-\frac{81}{4}=\frac{27}{4}\) (đpcm)

Dấu bằng xảy ra khi \(a=b=c=1\)

Ta có:

a³+b³+

Bạn tham khảo (hoàn toàn dùng Cô-si):

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