Question

[Nguyễn Nhật]

Cho a,b,c > 0 thỏa mãn : a+b+c = 1 . Tìm GTNN của biểu thức :

A = \(14\left(a^2+b^2+c^2\right)+\dfrac{ab+bc+ca}{a^2b+b^2c+c^2a}\)

Answer 1

[Akai Haruma]

Lời giải:

Xét

\((a+b+c)(a^2+b^2+c^2)=(a^3+b^3+c^3+ab^2+bc^2+ca^2)+a^2b+b^2c+c^2a\)

Áp dụng BĐT AM-GM:

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

\(\Leftrightarrow a^2b+b^2c+c^2a\leq \frac{a^2+b^2+c^2}{3}\) (do \(a+b+c=1\))

Do đó, \(A\geq 14(a^2+b^2+c^2)+\frac{3(ab+bc+ac)}{a^2+b^2+c^2}\)

\(\Leftrightarrow A\geq 14[(a+b+c)^2-2(ab+bc+ac)]+\frac{3(ab+bc+ac)}{(a+b+c)^2-2(ab+bc+ac)}\)

\(\Leftrightarrow A\geq 14-28(ab+bc+ac)+\frac{3(ab+bc+ac)}{1-2(ab+bc+ac)}\)

Đặt \(ab+bc+ac=t\)

Theo AM-GM thì \(ab+bc+ac\leq\frac{(a+b+c)^2}{3}\Leftrightarrow t\leq \frac{1}{3}\Rightarrow t\in (0,\frac{1}{3}]\)

Ta có: \(A\geq 14-28t+\frac{3t}{1-2t}\)

Ta sẽ cm rằng \(14-28t+\frac{3t}{1-2t}\geq \frac{23}{3}\Leftrightarrow \frac{14(1-2t)^2+3t}{1-2t}\geq \frac{23}{3}\)

\(\Leftrightarrow 168t^2-159t+42\geq 23-46t\)

\(\Leftrightarrow (3t-1)(56t-19)\geq 0\) \((\star)\)

Vì \(t\leq \frac{1}{3}\Rightarrow 3t-1,56t-19\leq 0\Rightarrow (3t-1)(56t-19)\geq 0\)

Do đó \((\star)\) đúng kéo theo \(14-28t+\frac{3t}{1-2t}\geq \frac{23}{3}\Rightarrow A\geq \frac{23}{3}\)

Vậy \(A_{\min}=\frac{23}{3}\Leftrightarrow a=b=c=\frac{1}{3}\)