\(P=\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\ge\frac{\left(1+1+2\right)^2}{a+b+c}=\frac{16}{4}=4\)
P=1/a+1/b+4/c > {1+1+2}^2/a+b+c
=16/4=16:4=4
Ta có:
\(P=\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\)
\(P=\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\ge\frac{\left(1+1+2\right)^2}{a+b+c}\)
\(P=\frac{16}{4}\)
\(P=4\)
Vậy \(P=4\)
\(P=\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\)
Theo BĐT SVAC
\(\Rightarrow P\ge\frac{\left(1+1+4\right)^2}{a+b+c}=\frac{36}{4}=9\)
Dấu "=" xảy ra khi .........
Ta có:
\(4P=4\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\ge\left(\frac{1}{\sqrt{a}}.\sqrt{a}+\frac{1}{\sqrt{b}}.\sqrt{b}+\frac{4}{\sqrt{c}}.\sqrt{c}\right)^2\)
\(=\left(1+1+2\right)^2=16\Rightarrow4P\ge16\Rightarrow P\ge4\)
Dấu bằng xảy ra khi a=b=1,c=2
Ta có: \(P=\frac{1}{a}+\frac{1}{b}+\frac{4}{c}=\frac{1^2}{a}+\frac{1^2}{b}+\frac{2^2}{c}\). Áp dụng BĐT SVac-xơ Ta có:
\(\frac{1^2}{a}+\frac{1^2}{b}+\frac{2^2}{c}\ge\frac{\left(1+1+2\right)^2}{a+b+c}=\frac{16}{4}=4\)
Vậy \(P_{min}=4\Leftrightarrow\frac{1}{a}=\frac{1}{b}=\frac{4}{c}\)
BĐT đó là TH đặc biệt của BĐT Bunyakovsky (Cauchy-Schwarz) mà sao lại ghi là BĐT SVac-xơ ???