Chủ đề này có 3 trả lời

[happypolla]

$Cho:\left\{\begin{matrix}a+b\leq 1 & & \\ a,b>0 & & \end{matrix}\right. Tim: min A=\frac{1}{a^2+b^2+1}+\frac{1}{2ab}$

[Hoang Nhat Tuan]

$Cho:\left\{\begin{matrix}a+b\leq 1 & & \\ a,b>0 & & \end{matrix}\right. Tim: min A=\frac{1}{a^2+b^2+1}+\frac{1}{2ab}$

Ta có:$A=\frac{1}{a^2+b^2+1}+\frac{1}{2ab+1}+(\frac{1}{2ab}-\frac{1}{2ab+1})\geq \frac{4}{(a+b)^2+2}+\frac{1}{2ab(2ab+1)}$

$\geq \frac{4}{3}+\frac{1}{2ab(2ab+1)}$

Lại có $1\geq a+b\geq 2\sqrt{ab}=>ab\leq \frac{1}{4}$

Do đó:$\frac{1}{2ab(2ab+1)}\geq \frac{4}{3}$

Vậy min $A=\frac{8}{3}$

[vda2000]

$Cho:\left\{\begin{matrix}a+b\leq 1 & & \\ a,b>0 & & \end{matrix}\right. Tim: min A=\frac{1}{a^2+b^2+1}+\frac{1}{2ab}$

Ta có: $A=\frac{1}{a^2+b^2+1}+\frac{1}{6ab}+\frac{1}{3ab}\geq\frac{4}{a^2+b^2+1+6ab}+\frac{1}{3ab}=\frac{4}{(a+b)^2+4ab+1}+\frac{1}{3ab}\geq\frac{4}{1+4.\frac{1}{4}+1}+\frac{1}{3.\frac{1}{4}}=\frac{4}{3}+\frac{4}{3}=\frac{8}{3}$

Do ta có: $a+b\leq 1$ nên cũng có: $ab\leq\frac{1}{4}$

[HoangVienDuy]

$Cho:\left\{\begin{matrix}a+b\leq 1 & & \\ a,b>0 & & \end{matrix}\right. Tim: min A=\frac{1}{a^2+b^2+1}+\frac{1}{2ab}$

áp dụng cauchy-schwarz ta có:

P=$\frac{1}{a^{2}+b^{2}+1}+\frac{\frac{1}{9}}{2ab}+\frac{\frac{8}{9}}{2ab}\geq \frac{(1+\frac{1}{3})^{2}}{(a+b)^{2}+1}+\frac{\frac{8}{9}}{2ab}\geq \frac{8}{3}$