Lời giải:
Ta có:
\(\text{VT}=2a-\frac{2ab^2}{a+b^2}+2b-\frac{2bc^2}{b+c^2}+2c-\frac{2ca^2}{c+a^2}\)
\(=2(a+b+c)-2\left(\frac{ab^2}{a+b^2}+\frac{bc^2}{b+c^2}+\frac{ca^2}{c+a^2}\right)(*)\)
Áp dụng BĐT AM-GM:
\(\frac{ab^2}{a+b^2}+\frac{bc^2}{b+c^2}+\frac{ca^2}{c+a^2}\leq \frac{ab^2}{2\sqrt{ab^2}}+\frac{bc^2}{2\sqrt{bc^2}}+\frac{ca^2}{2\sqrt{ca^2}}=\frac{\sqrt{ab^2}}{2}+\frac{\sqrt{bc^2}}{2}+\frac{\sqrt{ca^2}}{2}\)
\(\leq \frac{ab+b}{4}+\frac{bc+c}{4}+\frac{ca+a}{4}=\frac{ab+bc+ac+a+b+c}{4}(1)\)
Mà:
\((a+b+c)^2\geq 3(ab+bc+ac)=(a^2+b^2+c^2)(ab+bc+ac)\geq (ab+bc+ac)^2\)
\(\Rightarrow a+b+c\geq ab+bc+ac(2)\)
Từ \((1);(2)\Rightarrow \frac{ab^2}{a+b^2}+\frac{bc^2}{b+c^2}+\frac{ca^2}{c+a^2}\leq \frac{a+b+c}{2}(**)\)
Từ $(*); (**)\Rightarrow \text{VT}\geq a+b+c$ (đpcm)
Dấu "=" xảy ra khi $a=b=c=1$
CÁCH KHÁC:
Áp dụng BĐT Svarxo:
\(VT=2\left(\frac{a^2}{a+b^2}+\frac{b^2}{b+c^2}+\frac{c^2}{c+a^2}\right)\)\(\ge\frac{2\left(a+b+c\right)^2}{a+b+c+a^2+b^2+c^2}\)\(=\frac{2\left(a+b+c\right)^2}{a+b+c+3}\ge a+b+c\)
\(\Leftrightarrow2\left(a+b+c\right)^2-\left(a+b+c\right)\left(a+b+c+3\right)\ge0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a+b+c-3\right)\ge0\)
Đặt t=a+b+c(t>0)
\(\Rightarrow t\left(t-3\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}t\le0\\t\ge3\end{matrix}\right.\)
Giả sử t<3 hay a+b+c<3
=> Mỗi số a,b,c<3
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