Cho a,b,c > 0 và a+ b + c \(\le\dfrac{3}{2}\). Tìm Min của \(E=a+b+c+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
@Phùng Khánh Linh @Akai Haruma ...... giúp với
E = a + \(\dfrac{1}{4a}+b+\dfrac{1}{4b}+c+\dfrac{1}{4c}+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
áp dụng bdt cosi cho cac so duong co:
\(a+\dfrac{1}{4a}\ge2\sqrt{a.\dfrac{1}{4a}}\Leftrightarrow a+\dfrac{1}{4a}\ge1\)
\(b+\dfrac{1}{4b}\ge1,c+\dfrac{1}{4c}\ge1\)
dấu = xảy ra khi a=b=c = 1/2
CM: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{\dfrac{3}{2}}\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge6\)\(\Rightarrow\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{9}{2}\)\(\Rightarrow E\ge3+\dfrac{9}{2}\Rightarrow E\ge\dfrac{15}{2}\)
Vậy min E= 15/2 khi a=b=c=1/2
