Question

[Anh Phạm Xuân]

Câu hỏi:

a. Cho a,b,c\(\ge\)0. CMR: ( a+b )( b+c )( c+a )\(\ge\)8abc

b. Cho a+b\(\ge\)0 và a²+b²\(\le\)2. CMR: \(a\sqrt{3a\left(a+2b\right)}\)+\(b\sqrt{3b\left(b+2a\right)}\)\(\le\)6

c. Cho a,b,c\(\ge\)0 và (a+b)(b+c)(c+a) =1. CMR: ab+bc+ca\(\le\)\(\dfrac{3}{4}\)

Answer 1

[Tinh Lãm]

Trả lời:

a. Áp dụng BĐT Cô-si: x + y\(\ge\) \(2\sqrt{xy}\) (với x,y\(\ge\)0)

Ta có: a + b\(\ge\)\(2\sqrt{ab}\)

b+c\(\ge\)\(2\sqrt{bc}\)

c+a\(\ge\)\(2\sqrt{ca}\)

\(\Rightarrow\) (a+b)(b+c)(c+a) \(\ge\)\(8\sqrt{a^2b^2c^2}\)= 8abc (đpcm)

b. Áp dụng BĐT Cô-si: \(\sqrt{ab}\)\(\le\)\(\dfrac{a+b}{2}\) ( với a,b\(\ge\)0)

Ta có: \(\sqrt{3a\left(a+2b\right)}\)\(\le\)\(\dfrac{3a+a+2b}{2}\)=\(\dfrac{4a+2b}{2}\)=2a+b

\(\Rightarrow\) \(a\sqrt{3a\left(a+2b\right)}\)\(\le\)a(2a+b) = 2a²+ab

CMTT: \(b\sqrt{3b\left(b+2a\right)}\)\(\le\)b(2b+a) = 2b²+ab

\(\rightarrow\)\(a\sqrt{3a\left(a+2b\right)}\)+\(b\sqrt{3b\left(2b+a\right)}\)\(\le\) 2a²+ab+2b²+ab

= 2(a²+b²)+2ab =6(đpcm)

c. Áp dụng BĐT Cô-si với 3 số a+b; b+c;c+a

Ta có: (a+b)(b+c)(c+a)\(\le\)\(\left(\dfrac{2\left(a+b+c\right)}{3}\right)^3\)

\(\Leftrightarrow\) 1 \(\le\) \(\dfrac{8}{27}\left(a+b+c\right)^3\)

\(\Leftrightarrow\) (a+b+c)³ \(\ge\) \(\dfrac{8}{27}\)

\(\Leftrightarrow\) a+b+c \(\ge\) \(\dfrac{3}{2}\) (1)

Lại có: (a+b)(b+c)(c+a) = (a+b+c)(ab+bc+ca) -abc

\(\Leftrightarrow\) 1= (a+b+c)(ab+bc+ca) - abc

\(\Leftrightarrow\) ab+bc+ca = \(\dfrac{1+abc}{a+b+c}\) (2)

Theo câu a. (a+b)(b+c)(c+a) \(\ge\) 8abc

\(\Leftrightarrow\) 1 \(\ge\) 8abc

\(\Leftrightarrow\) abc \(\le\)\(\dfrac{1}{8}\) (3)

Từ (1),(3) kết hợp với (2)

\(\Rightarrow\) ab+bc+ca \(\le\) \(\dfrac{1+\dfrac{1}{8}}{\dfrac{3}{2}}\) = \(\dfrac{3}{4}\) (đpcm)