Em tham khảo nha:
P=\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{a+1}{a}.\frac{b+1}{b}.\frac{c+1}{c}\)
P\(=\frac{\left(a+1\right)\left(b+1\right)\left(c+1\right)}{abc}\)
P\(=\frac{abc+ab+bc+ca+a+b+c+1}{abc}\)
P\(=1+\frac{ab+bc+ca}{abc}+\frac{2}{abc}\)(1)
( vì a+b+c=1 nên a+b+c+1=2)
Áp dụng BĐT Cô-si ta có:
\(ab+bc+ca\ge3\sqrt[3]{ab.bc.ca}=3\sqrt[3]{a^2b^2c^2}\)
\(\)Tương tự:
\(a+b+c\ge3\sqrt[3]{abc}\)
\(\Rightarrow\sqrt[3]{abc}\le\frac{a+b+c}{3}\)
\(\Rightarrow abc\le\left(\frac{a+b+c}{3}\right)^3\)
\(P=\left(1\right)\ge1+\frac{3\sqrt[3]{a^2b^2c^2}}{abc}+\frac{2}{\left(\frac{a+b+c}{3}\right)^3}=1+\frac{3}{\sqrt[3]{abc}}+\frac{2}{\frac{1}{27}}\)
( \(\frac{\sqrt[3]{a^2b^2c^2}}{abc}=\sqrt[3]{\frac{a^2b^2c^2}{a^3b^3c^3}}=\sqrt[3]{\frac{1}{abc}}=\frac{1}{\sqrt[3]{abc}}\))
P\(\ge1+54+\frac{9}{3\sqrt[3]{abc}}\)
P\(\ge55+\frac{9}{a+b+c}=55+\frac{9}{1}=64\)
Vậy GTNN của P là P=64 . Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
Cách 2:
Áp dụng BĐT Holder và BĐT AM-GM:
\(P\ge\left(1+\frac{1}{\sqrt[3]{abc}}\right)^3\ge\left(1+\frac{1}{\frac{a+b+c}{3}}\right)^3=64\)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)
Cách 3:
Áp dụng BĐT AM -GM (Cô si):
\(P=\left(1+\frac{1}{3a}+\frac{1}{3a}+\frac{1}{3a}\right)\left(1+\frac{1}{3b}+\frac{1}{3b}+\frac{1}{3b}\right)\left(1+\frac{1}{3c}+\frac{1}{3c}+\frac{1}{3c}\right)\)
\(\ge64\sqrt[4]{\frac{1}{\left(27abc\right)^3}}\ge64\sqrt[4]{\frac{1}{\left(a+b+c\right)^9}}=64\)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)
Hồ Bảo TrâmTrần Thanh PhươngNguyễn Thành Trươngbuithianhtho
Akai Haruma
