Lời giải:
Xét hiệu:
\(2(a^4+c^4)-(a^3+c^3)(a+c)=2(a^4+c^4)-(a^4+a^3c+ac^3+c^4)\)
\(=a^4+c^4-a^3c-ac^3=(a-c)(a^3-c^3)=(a-c)^2(a^2+ac+c^2)\geq 0\)
với mọi \(a,c>0\)
Do đó: \(2(a^4+c^4)\geq (a^3+c^3)(a+c)\Leftrightarrow \frac{a^4+c^4}{a^3+c^3}\geq \frac{a+b}{2}\)
Hoàn toàn tương tự ta có:
\(\left\{\begin{matrix}
\frac{b^4+c^4}{b^3+c^3}\geq \frac{b+c}{2}\\
\frac{a^4+b^4}{a^3+b^3}\geq \frac{a+b}{2}\end{matrix}\right.\)
Cộng theo vế các BĐT thu được:
\(\frac{a^4+b^4}{a^3+b^3}+\frac{b^4+c^4}{b^3+c^3}+\frac{c^4+a^4}{c^3+a^3}\geq \frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}=a+b+c=2018\)
Ta có đpcm.
Dấu bằng xảy ra khi $a=b=c=\frac{2018}{3}$
\(\dfrac{a^4+b^4}{a^3+b^3}+\dfrac{b^4+c^4}{b^3+c^3}+\dfrac{c^4+a^4}{c^3+a^3}\ge2018\)
\(\Leftrightarrow\dfrac{a^4+b^4}{a^3+b^3}+\dfrac{b^4+c^4}{b^3+c^3}+\dfrac{c^4+a^4}{c^3+a^3}\ge a+b+c\)
\(\LeftrightarrowΣ_{cyc}\dfrac{a^3\left(a-c\right)+b^3\left(b-c\right)}{a^3+b^3}\ge0\)
\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)\left(\dfrac{a^3}{c^3+a^3}-\dfrac{b^3}{b^3+c^3}\right)\right)\ge0\)
\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\dfrac{c^3\left(a^2+ab+b^2\right)}{\left(a+c\right)\left(a^2-ac+c^2\right)\left(b+c\right)\left(b^2-bc+c^2\right)}\right)\ge0\)
Dễ thấy BĐT cuối luôn đúng nên ta có ĐPCM
Dấu "=" <=> \(a=b=c=\dfrac{2018}{3}\)
