Chủ đề này có 2 trả lời

[Zimmi]

$\left\{\begin{matrix} a,b,c,d >0 & & \\ a+b+c+d=1& & \end{matrix}\right. CMR (1+\frac{1}{a})(1+\frac{1}{b}(1+\frac{1}{c})(1+\frac{1}{d})\geq 54$

Bài viết đã được chỉnh sửa nội dung bởi Zimmi: 03-10-2013 - 22:11

[Hoang Tung 126]

Kia phải là $\geq 5^4$ chứ.Theo bđt Cosi cho 5 số ta có :$1+\frac{1}{a}=1+\frac{1}{4a}+\frac{1}{4a}+\frac{1}{4a}+\frac{1}{4a}\geq 5\sqrt[5]{\frac{1}{4^4.a^4}}.$

Tương tự :$1+\frac{1}{b}\geq \frac{5}{\sqrt[5]{4^4.b^4}},1+\frac{1}{c}\geq \frac{5}{\sqrt[5]{4^4.c^4}},1+\frac{1}{d}\geq \frac{5}{\sqrt[5]{d^4.4^4}}= > (1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})(1+\frac{1}{d})\geq \frac{5^4}{\sqrt[5]{4^{16}.(abcd)^4}}$(1)

Mặt khác :$abcd\leq \frac{(a+b+c+d)^4}{4^4}=\frac{1}{256}= > (abcd)^4\leq \frac{1}{256^4}= > \sqrt[5]{(abcd)^4}\leq \sqrt[5]{\frac{1}{256^4}}= > \sqrt[5]{4^{16}.(abcd)^4}\leq \sqrt[5]{\frac{4^{16}}{256^4}}=1$(2)

Từ (1),(2) $= > A\geq 5^4$(đpcm)

Dấu = xảy ra khi a=b=c=d=$\frac{1}{4}$

[Trang Luong]

$\left\{\begin{matrix} a,b,c,d >0 & & \\ a+b+c+d=1& & \end{matrix}\right. CMR (1+\frac{1}{a})(1+\frac{1}{b}(1+\frac{1}{c})(1+\frac{1}{d})\geq 54$

Ta có $a+b+c+d=1$

$\Rightarrow \prod \left ( 1+\frac{1}{a} \right )=\prod \left ( 1+ \frac{a+b+c+d}{a} \right )=\prod \left ( 2+\frac{b}{a}+\frac{c}{a}+\frac{d}{a} \right )=\prod \left ( \frac{b}{a}+\frac{c}{a}+\frac{d}{a}+1+1 \right )\geq \prod \left ( 5\sqrt[5]{\frac{bcd}{a^3}} \right )=5^{4}.\sqrt[5]{\frac{(abcd)^3}{(abcd)^3}}=5^4$