Từ a+b+c=6 \(\Rightarrow\)a+b=6-c

Ta có: ab+bc+ac=9\(\Leftrightarrow\)ab+c(a+b)=9

                               \(\Leftrightarrow\)ab=9-c(a+b)

           Mà a+b=6-c (cmt)

                                \(\Rightarrow\)ab=9-c(6-c)

                                \(\Rightarrow\)ab=9-6c+c²

Ta có: (b-a)²\(\ge\)0 \(\forall\)b, c

  \(\Rightarrow\)b²+a²-2ab\(\ge\)0

  \(\Rightarrow\)(b+a)²-4ab\(\ge\)0

  \(\Rightarrow\)(a+b)²\(\ge\)4ab

Mà a+b=6-c (cmt)

         ab= 9-6c+c² (cmt)

  \(\Rightarrow\)(6-c)²\(\ge\)4(9-6c+c²)

  \(\Rightarrow\)36+c²-12c\(\ge\)36-24c+4c²

  \(\Rightarrow\)36+c²-12c-36+24c-4c²\(\ge\)0

  \(\Rightarrow\)-3c²+12c\(\ge\)0

  \(\Rightarrow\)3c²-12c\(\le\)0

  \(\Rightarrow\)3c(c-4)\(\le\)0

  \(\Rightarrow\)c(c-4)\(\le\)0

\(\Rightarrow\hept{\begin{cases}c\ge0\\c-4\le0\end{cases}}\)hoặc\(\hept{\begin{cases}c\le0\\c-4\ge0\end{cases}}\)

*\(\hept{\begin{cases}c\ge0\\c-4\le0\end{cases}\Leftrightarrow\hept{\begin{cases}c\ge0\\c\le4\end{cases}\Leftrightarrow}0\le c\le4}\)

*

28 nha bạn

we had abc+(4-a)(4-b)(4-c)\(\ge0\). khai triển ta có \(ab+bc+ca\ge8\)( maybe)

\(P=\left(a+b+c\right)^2-\left(ab+bc+ca\right)\le6^2-8=28\)

Dấu = xảy ra (a,b,c)~(0;2;4) và các hoán vị

Bạn tham khảo ở đây nhé

https://olm.vn/hoi-dap/detail/49527613309.html

Ta có: \(a^2+b^2+c^2\ge ab+bc+ca\ge\sqrt[]{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

Do đó:

\(VT\le\dfrac{2a^3}{2\sqrt{a^6bc}}+\dfrac{2b^3}{2\sqrt{b^6ac}}+\dfrac{2c^3}{2\sqrt{c^3ab}}=\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{abc}}=\dfrac{\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}{abc}\)

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

Dấu "=" xảy ra khi \(a=b=c=1\)

Vô lí vì a+b+c=0\(\Rightarrow\frac{5}{a+b+c}\)không có đáp án