Câu hỏi của Trọng Chi Ca Vâu

Question

cho a,b,c là các số thực dương.CMR:$\dfrac{a^5}{a^2+ab+b^2}+\dfrac{b^5}{b^2+bc+c^2}+\dfrac{c^5}{c^2+ca+a^2}\ge\dfrac{a^3+b^3+c^3}{3}$

Akai Haruma · Accepted Answer

Lời giải:

Áp dụng BĐT Cauchy_ Schwarz ta có:

$\text{VT}=\frac{a^6}{a^3+a^2b+ab^2}+\frac{b^6}{b^3+b^2c+bc^2}+\frac{c^6}{c^3+c^2a+ca^2}$

$\geq \frac{(a^3+b^3+c^3)^2}{a^3+a^2b+ab^2+b^3+b^2c+bc^2+c^3+c^2a+ca^2}$

$\Leftrightarrow \text{VT}\geq \frac{(a^3+b^3+c^3)^2}{a^3+b^3+c^3+ab(a+b)+bc(b+c)+ac(a+c)}$ (I)

Áp dụng BĐT Am-Gm ta có:

$\left\{\begin{matrix}
a^3+a^3+b^3\geq 3a^2b\ 
b^3+b^3+c^3\geq 3b^2c\

c^3+c^3+a^3\geq 3c^2a\end{matrix}\right.\Rightarrow 3(a^3+b^3+c^3)\geq 3(a^2b+b^2c+c^2a)$

$\Leftrightarrow a^3+b^3+c^3\geq a^2b+b^2c+c^2a$ (1)

Tương tự:

$\left\{\begin{matrix}
a^3+b^3+b^3\geq 3ab^2\ 
b^3+c^3+c^3\geq 3bc^2\

c^3+a^3+a^3\geq 3ca^2\end{matrix}\right.\Rightarrow 3(a^3+b^3+c^3)\geq 3(ab^2+bc^2+ca^2)$

$\Leftrightarrow a^3+b^3+c^3\geq ab^2+bc^2+ca^2(2)$

Từ $(1);(2)\Rightarrow 2(a^3+b^3+c^3)\geq ab(a+b)+bc(b+c)+ac(c+a)$

$\Rightarrow a^3+b^3+c^3+ab(a+b)+bc(b+c)+ac(c+a)\leq 3(a^3+b^3+c^3)$ (II)

Từ $(I);(II)\Rightarrow \text{VT}\geq \frac{(a^3+b^3+c^3)^2}{a^3+b^3+c^3+ab(a+b)+bc(b+c)+ac(a+c)}\geq \frac{(a^3+b^3+c^3)^2}{3(a^3+b^3+c^3)}$

$\Leftrightarrow \text{VT}\geq \frac{a^3+b^3+c^3}{3}$ (đpcm)

Dấu bằng xảy ra khi $a=b=c$Câu trả lời: Lời giải: