Chủ đề này có 4 trả lời

[daotuanminh]

Cho a,b,c là 3 cạnh tam giác. Tìm Min:

E=$\frac{4a}{b+c-a}+\frac{9b}{a+c-b}+\frac{16c}{a+b-c}$

[Trang Luong]

Cho a,b,c là 3 cạnh tam giác. Tìm Min:

E=$\frac{4a}{b+c-a}+\frac{9b}{a+c-b}+\frac{16c}{a+b-c}$

http://diendantoanho...c-bfrac16cab-c/

[kimchitwinkle]

Đặt $\left\{\begin{matrix} b+c-a=x & \\ a+c-b=y& \\ a+b-c=z & \end{matrix}\right.$

 => x,y,z >0 ( theo bđt tam giác )

      $\left\{\begin{matrix} a=\frac{y+z}{2} & \\ b=\frac{z+x}{2} & \\ c=\frac{x+y}{2} & \end{matrix}\right.$

Khi đó : 

 E = $\frac{4\left ( y+z \right )}{2x}+\frac{9\left ( x+z \right )}{2y}+\frac{16\left ( x+y \right )}{2z}$

 E = $\frac{4y+4z}{2x}+\frac{9x+9z}{2y}+\frac{16x+16y}{2z}$

 E = $\frac{4y}{2x}+\frac{4z}{2x}+\frac{9x}{2y}+\frac{9z}{2y}+\frac{16x}{2z}+\frac{16y}{2z}$

AD AM - GM cho  $\frac{4y}{2x} ; \frac{9x}{2y}$ có:

  $\frac{4y}{2x}+\frac{9x}{2y}\geq 2\sqrt{\frac{4y}{2x}.\frac{9x}{2y}}=6$

Tương tự : $\frac{4z}{2x}+\frac{16x}{2z}\geq 8$

                  $\frac{9z}{2y}+\frac{16y}{2z}\geq 12$

Cộng từng vế => E $\geq 6+8+12 = 26$ ( ko đôỉ)

Dấu = <=> $\left\{\begin{matrix} \frac{4y}{2x}=\frac{9x}{2y} & & & \\ \frac{4z}{2x}=\frac{16x}{2z}& & & \\ \frac{9z}{2y}=\frac{16y}{2z}& & & \end{matrix}\right.$

 <=> $\left\{\begin{matrix} x=2 & \\ y=3 & \\ z=4 & \end{matrix}\right.$

<=> $\left\{\begin{matrix} a=\frac{7}{2} & \\ b=3 & \\ c=\frac{5}{2}& \end{matrix}\right.$

Vậy min E = 26 <=> $\left\{\begin{matrix} a=\frac{7}2{} & \\ b=3& \\ c= \frac{5}{2}& \end{matrix}\right.$

[daotuanminh]

Cách bạn giống cách tớ

[KietLW9]

Cho a,b,c là 3 cạnh tam giác. Tìm Min:

E=$\frac{4a}{b+c-a}+\frac{9b}{a+c-b}+\frac{16c}{a+b-c}$

Ta có: $E=\frac{4a}{b+c-a}+\frac{9b}{c+a-b}+\frac{16c}{a+b-c}=4(\frac{a}{b+c-a}+\frac{1}{2})+9(\frac{b}{c+a-b}+\frac{1}{2})+16(\frac{c}{a+b-c}+\frac{1}{2})-\frac{29}{2}=\frac{a+b+c}{2}(\frac{4}{b+c-a}+\frac{9}{c+a-b}+\frac{16}{a+b-c})-\frac{29}{2}\geqslant \frac{a+b+c}{2}.\frac{(2+3+4)^2}{a+b+c}-\frac{29}{2}=26$