Thấy : \(a;b;c\ge0;a+b+c=1\)  \(\Rightarrow1-a;1-b;1-c\ge0\)

AD BĐT AM - GM ta được :  \(4\left(1-a\right)\left(1-c\right)\le\left(2-a-c\right)^2=\left[2-\left(1-b\right)\right]^2=\left(b+1\right)^2\)

\(\Rightarrow4\left(1-a\right)\left(1-b\right)\left(1-c\right)\le\left(1-b\right)\left(b+1\right)^2=\left(1-b^2\right)\left(b+1\right)\le1.\left(b+1\right)=b+1=b+\left(a+b+c\right)=a+2b+c\)

( đpcm ) 

Xét \(VT=a+2b+c=1+b\left(1\right)\)

Áp dụng BĐT AG-GM:

\(4\left(1-a\right)\left(1-c\right)\le\left(1-a+1-c\right)^2=\left(2-a-c\right)^2=\left(1+a+b+c-a-c\right)^2=\left(1+b\right)^2\left(2\right)\)

\(\Rightarrow4\left(1-a\right)\left(1-b\right)\left(1-c\right)\le\left(1-b\right)\left(1+b\right)^2\)

Mà \(\left(1-b\right)\left(1+b\right)^2-\left(1-b\right)=\left(1+b\right)\left(1-b^2-1\right)=-b^2\left(1+b\right)\le0,\forall b\ge0\)

Do đó \(\left(1-b\right)\left(1+b\right)^2\le1+b\left(3\right)\)

Từ \(\left(1\right)\left(2\right)\left(3\right)\) ta có ĐPCM

Dấu "=" \(\Leftrightarrow a=c=\dfrac{1}{2};b=0\) 

Ta có:

Dấu “=” xảy ra khi và chỉ khi

Vậy số bộ a,b,c thỏa mãn điều kiện đã cho là 1.

Chọn B.

Vì a, b, c không âm và có tổng bằng 1 nên  0 ≤ a , b , c ≤ 1 ⇒ a ( 1 − a ) ≥ 0 b ( 1 − b ) ≥ 0 c ( 1 − c ) ≥ 0 ⇒ a ≥ a 2 b ≥ b 2 c ≥ c 2 ⇒ 5 a + 4 ≥ a 2 + 4 a + 4 = ( a + 2 ) 2 = a + 2 T ư ơ n g   t ự :   5 b + 4 ≥ b + 2 ; 5 c + 4 ≥ c + 2 ⇒ 5 a + 4 + 5 b + 4 + 5 c + 4 ≥ ( a + b + c ) + 6 = 7   ( đ p c m )

Áp dụng BĐT BSC:

\(P=\dfrac{1}{a}+\dfrac{4}{b}+\dfrac{9}{c}\ge\dfrac{\left(1+2+3\right)^2}{a+b+c}=\dfrac{36}{1}=36\)

\(minP=36\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{6}\\b=\dfrac{1}{3}\\c=\dfrac{1}{2}\end{matrix}\right.\)

Ta có \(\left(a+b+c\right)\left(ab+bc+ca\right)=\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)

Mà \(abc\le\frac{1}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\) (AM-GM)

\(\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\le\left(a+b\right)\left(b+c\right)\left(c+a\right)+\frac{1}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(\Rightarrow\frac{9}{8}\ge\left(a+b+c\right)\left(ab+bc+ca\right)\ge\sqrt{3\left(ab+bc+ca\right)}.\left(ab+bc+ca\right)\)

\(\Rightarrow3\left(ab+bc+ca\right)^3\le\frac{81}{64}\)

\(\Rightarrow ab+bc+ca\le\frac{3}{4}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{2}\)

Ta có: \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=1\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ca\right)-abc=1\)

Áp dụng BĐT Cô si ta có

\(1=\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}\)\(=8abc\)

\(\Rightarrow abc\le\frac{1}{8}\)

mặt khác: \(1=\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\left(\frac{2a+2b+2c}{3}\right)^3\)

\(\Rightarrow a+b+c\ge\frac{3}{2}\)

\(\Rightarrow ab+bc+ca=\frac{1+abc}{a+b+c}\le\frac{1+\frac{1}{8}}{\frac{3}{2}}=\frac{3}{4}\)

\(VT\leΣ\frac{1}{a^2+b^2+1}\le\frac{a^2+b^2+c^2+6}{\left(a+b+c\right)^2}\le\frac{\left(Σa\right)^2}{\left(Σa\right)^2}=1=VP\)

Bạn giải rõ ra được không

\(VT=\Sigma\frac{1}{\frac{a^3}{b}+\frac{b^3}{a}+1}=\Sigma\frac{1}{\frac{a^4}{ab}+\frac{b^4}{ab}+1}\)

Áp dụng BĐT Cauchy-schwar ta có:

\(VT\le\Sigma\frac{1}{\frac{\left(a^2+b^2\right)^2}{2ab}+1}\le\Sigma\frac{1}{\frac{\left(a^2+b^2\right).2ab}{2ab}+1}=\Sigma\frac{1}{a^2+b^2+1}\)\(=\Sigma\frac{c^2+2}{\left(c^2+2\right)\left(a^2+b^2+1\right)}=\Sigma\frac{c^2+2}{\left(a^2c^2+1\right)+\left(b^2c^2+1\right)+\left(a^2+b^2\right)+a^2+b^2+c^2}=\Sigma\frac{c^2+2}{\left(a+b+c\right)^2}=\Sigma\frac{a^2+b^2+c^2+6}{\left(a+b+c\right)^2}\)Áp dụng BĐT AM-GM ta có:

\(ab+bc+ca+ab+bc+ca\ge6.\sqrt[6]{a^4b^4c^4}=6\)

\(\Rightarrow\)\(VT\le\frac{a^2+b^2+c^2+2ab+2bc+2ca}{\left(a+b+c\right)^2}=\frac{\left(\Sigma a\right)^2}{\left(\Sigma a\right)^2}=1\)

Dấu ' = " xảy ra <=> a=b=c

đpcm

Nhân vế với vế của giả thiết:

\(\left(a+\dfrac{1}{b}\right)\left(b+\dfrac{1}{c}\right)\left(c+\dfrac{1}{a}\right)=\dfrac{28}{3}\)

\(\Leftrightarrow\left(ab+\dfrac{1}{bc}+\dfrac{a}{c}+1\right)\left(c+\dfrac{1}{a}\right)=\dfrac{28}{3}\)

\(\Leftrightarrow abc+\dfrac{1}{abc}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+a+b+c=\dfrac{28}{3}\) (1)

Cộng vế với vế giả thiết:

\(\Rightarrow a+b+c+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=4+1+\dfrac{7}{3}=\dfrac{22}{3}\) (2)

(1);(2) \(\Rightarrow abc+\dfrac{1}{abc}+\dfrac{22}{3}=\dfrac{28}{3}\)

\(\Rightarrow abc+\dfrac{1}{abc}=2\)

\(\Rightarrow\left(abc\right)^2-2\left(abc\right)+1=0\)

\(\Rightarrow\left(abc-1\right)^2=0\)

\(\Rightarrow abc=1\)