Chủ đề này có 3 trả lời

[leanh9adst]

Cho a,b,c > 0 thỏa mãn $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 2$

Tìm GTLN của P =abc

[HoangAnhKhoaKing]

Ta có: $\frac{1}{a+1}\geq 2-(\frac{1}{b+1}+\frac{1}{c+1})=(1-\frac{1}{b+1})+(1-\frac{1}{c+1})=\frac{b}{b+1}+\frac{c}{c+1}\geq$$2\times \sqrt{\frac{bc}{(b+1)(c+1)}}$ (1)

CMTT: $\frac{1}{b+1}\geq 2\times \sqrt{\frac{ac}{(a+1)(c+1)}}$ (2)

$\frac{1}{c+1}\geq 2\times \sqrt{\frac{ab}{(a+1)(b+1)}}$ (3)

Lấy (1)$\times$(2)$\times$(3) , ta được: $\frac{1}{(a+1)(b+1)(c+1)}\geq \frac{8abc}{(a+1)(b+1)(c+1)}$

$\Leftrightarrow 8abc\leq 1\Leftrightarrow abc\leq \frac{1}{8}$

Dấu "=" xr $\Leftrightarrow a=b=c=\frac{1}{2}$

$\Rightarrow max P=\frac{1}{8}\Leftrightarrow a=b=c=\frac{1}{2}$

[haichau0401]

Cho a,b,c > 0 thỏa mãn $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 2$

Tìm GTLN của P =abc

Ta có:

$\frac{1}{a+1}\geq 1-\frac{1}{b+1}+1-\frac{1}{c+1}=\frac{b}{b+1}+\frac{c}{c+1}\geq 2\sqrt{\frac{bc}{(b+1)(c+1 )}}$

Đến đây tương tự nhân vế theo vế là ra!

[minhhien2001]

$\frac{a+1-a}{a+1}+\frac{b+1-b}{b+1}+\frac{c+1-c}{c+1}\geqslant 2\Leftrightarrow \frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\leqslant 1$.
áp dụng BĐT bunhiacoski ta có $(a+b+c+3)(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1})\geqslant (\sqrt{a}+\sqrt{b}+\sqrt{c})^2\Rightarrow (a+b+c+3)\geqslant (\sqrt{a}+\sqrt{b}+\sqrt{c})^2 \Leftrightarrow \sqrt{ab}+\sqrt{ac}+\sqrt{bc} \leqslant \frac{3}{2}$.
=> $\frac{3}{2}\geqslant \sqrt{ab}+\sqrt{ac}+\sqrt{bc}\geqslant 3\sqrt[3]{abc}=>\frac{1}{8}\geqslant abc$
dấu "=" xảy ra khi a=b=c$= \frac{1}{2}$