Cho các số dương a,b,c tm:
a+b+c=1. Tìm Max M=\(\sqrt{a^2+abc}+\sqrt{b^2+abc}+\sqrt{c^2+abc}+9\sqrt{abc}\)
Lời giải:
Ta có:
\(M=\sqrt{a^2+abc}+\sqrt{b^2+abc}+\sqrt{c^2+abc}+9\sqrt{abc}\)
\(M=\sqrt{a(a+bc)}+\sqrt{b(b+ac)}+\sqrt{c(c+ab)}+9\sqrt{abc}\)
Áp dụng BĐT Bunhiacopxky:
\([\sqrt{a(a+bc)}+\sqrt{b(b+ac)}+\sqrt{c(c+ab)}]^2\leq (a+b+c)(a+bc+b+ac+c+ab)\)
\(\Leftrightarrow \sqrt{a(a+bc)}+\sqrt{b(b+ac)}+\sqrt{c(c+ab)}\leq \sqrt{1+ab+bc+ac}\)
Theo hệ quả của BĐT AM-GM: \(ab+bc+ac\leq \frac{(a+b+c)^2}{3}=\frac{1}{3}\)
\(\Rightarrow \sqrt{a(a+bc)}+\sqrt{b(b+ac)}+\sqrt{c(c+ab)}\leq \frac{2\sqrt{3}}{3}(1)\)
AM-GM: \(a+b+c\geq 3\sqrt[3]{abc}\Rightarrow abc\leq \frac{1}{27}\Rightarrow 9\sqrt{abc}\leq \sqrt{3}(2)\)
Từ (1);(2) suy ra: \(M\leq \frac{2\sqrt{3}}{3}+\sqrt{3}=\frac{5\sqrt{3}}{3}\)
Vậy \(M_{\max}=\frac{5\sqrt{3}}{3}\) . Dấu bằng xảy ra khi \(a=b=c=\frac{1}{3}\)
