Cho a,b,c,d là các số thực thoả mãn điều kiện
\(abc+bcd+cda+dab=a+b+c+d+\sqrt{2012}\)
CMR: \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\left(d^2+1\right)\ge2012\)
K
Khách
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Những câu hỏi liên quan
12 tháng 12 2017
Ta có:
\(\sqrt{2012}=abc+bcd+cda+dab-a-b-c-d=\left(bc-1\right)\left(a+d\right)+\left(ad-1\right)\left(b+c\right)\)
\(\Leftrightarrow2012=\left[\left(bc-1\right)\left(a+d\right)+\left(ad-1\right)\left(b+c\right)\right]^2\)
\(\le\left[\left(bc-1\right)^2+\left(b+c\right)^2\right]\left[\left(ad-1\right)^2+\left(a+d\right)^2\right]\)
\(=\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\left(d^2+1\right)\)
LM
12 tháng 12 2017
\(GT\Leftrightarrow2012=\left[\left(bc-1\right)\left(a+d\right)+\left(a+c\right)\left(ad-1\right)\right]^2\le\left[\left(bc-1\right)^2+\left(b+c^2\right)\right]\)
\(\left[\left(ad-1\right)^2+\left(a+d\right)^2\right]=\left(b^2+1\right)\left(c^2+1\right)\left(a^2+1\right)\left(d^2+1\right)\)
P/s: Mình không chắc đâu ! Tham khảo nha!
UK
9 tháng 10 2017
https://diendantoanhoc.net/topic/76281-bdt-thi-h%E1%BB%8Dc-sinh-gi%E1%BB%8Fi-t%E1%BB%89nh-l%E1%BB%9Bp-9-nam-2011-2012/
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 4 2021
Theo nguyên lý Dirichlet, trong 3 số a;b;c luôn có ít nhất 2 số cùng phía so với 1
Không mất tính tổng quát, giả sử đó là a và b
\(\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\)
\(\Leftrightarrow ab+1\ge a+b\)
\(\Leftrightarrow2\left(ab+1\right)\ge\left(a+1\right)\left(b+1\right)\)
\(\Rightarrow\dfrac{2}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{2}{2\left(ab+1\right)\left(c+1\right)}=\dfrac{1}{\left(ab+1\right)\left(c+1\right)}=\dfrac{1}{\left(\dfrac{1}{c}+1\right)\left(c+1\right)}=\dfrac{c}{\left(c+1\right)^2}\)
Lại có:
\(\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{a}{b}}+1.1\right)^2}+\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{b}{a}}+1\right)^2}\ge\dfrac{1}{\left(ab+1\right)\left(\dfrac{a}{b}+1\right)}+\dfrac{1}{\left(ab+1\right)\left(\dfrac{b}{a}+1\right)}=\dfrac{1}{ab+1}\)
\(\Rightarrow P\ge\dfrac{1}{ab+1}+\dfrac{1}{\left(c+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}=\dfrac{1}{\dfrac{1}{c}+1}+\dfrac{1}{\left(c+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}\)
\(\Rightarrow P\ge\dfrac{c}{c+1}+\dfrac{c+1}{\left(c+1\right)^2}=\dfrac{c\left(c+1\right)+c+1}{\left(c+1\right)^2}=\dfrac{\left(c+1\right)^2}{\left(c+1\right)^2}=1\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
11 tháng 4 2017
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\)
Ta có:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{8}+\frac{1+c}{8}\ge\frac{3a}{4}\)
\(\Leftrightarrow\frac{a^3}{\left(1+b\right)\left(1+c\right)}\ge\frac{6a-b-c-2}{8}\)
Tương tự ta có: \(\hept{\begin{cases}\frac{b^3}{\left(1+c\right)\left(1+a\right)}\ge\frac{6b-c-a-2}{8}\\\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{6c-a-b-2}{8}\end{cases}}\)
Cộng vế theo vế ta được
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{6a-b-c-2}{8}+\frac{6b-c-a-2}{8}+\frac{6c-a-b-2}{8}\)
\(=\frac{a+b+c}{2}-\frac{3}{4}\ge\frac{3}{2}.\sqrt[3]{abc}-\frac{3}{4}=\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\)
\(\sqrt{2012}=\left(abc+bcd-a-d\right)+\left(cda+dab-c-b\right)\)
\(=\left(bc-1\right)\left(a+d\right)+\left(c+b\right)\left(ad-1\right)\)
\(\Rightarrow2012=\left[\left(bc-1\right)\left(a+d\right)+\left(c+b\right)\left(ad-1\right)\right]^2\)
\(\le\left[\left(bc-1\right)^2+\left(c+b\right)^2\right]\left[\left(a+d\right)^2+\left(ad-1\right)^2\right]\)
\(=\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\left(d^2+1\right)\)