Cho M = \(\dfrac{a}{a+b+c}+\dfrac{b}{a+b+d}+\dfrac{c}{b+c+d}+\dfrac{d}{a+d+c}\) (a,b,c thuộc N sao)
CMR: M ko là số tự nhiên
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Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+d}+\dfrac{d^2}{a+d}\)
\(\ge\dfrac{\left(a+b+c+d\right)^2}{a+b+b+c+c+d+d+a}\)
\(=\dfrac{\left(a+b+c+d\right)^2}{2\left(a+b+c+d\right)}=\dfrac{a+b+c+d}{2}=\dfrac{1}{2}=VP\)
Đẳng thức xảy ra khi \(a=b=c=d=\dfrac{1}{4}\)
Bài 1:
\(3^{-1}.3^n+4.3^n=13.3^5\)
\(\Rightarrow3^{n-1}+4.3.3^{n-1}=13.3^5\)
\(\Rightarrow3^{n-1}\left(1+4.3\right)=13.3^5\)
\(\Rightarrow3^{n-1}.13=13.3^5\)
\(\Rightarrow3^{n-1}=3^5\)
\(\Rightarrow n-1=5\)
\(\Rightarrow n=6\)
Vậy n = 6
Bài 2a: Câu hỏi của Nguyễn Trọng Phúc - Toán lớp 7 | Học trực tuyến
b) Ta có:
\(\dfrac{1^2}{a}+\dfrac{1^2}{b}+\dfrac{1^2}{c}+\dfrac{1^2}{d}\ge\dfrac{\left(1+1+1+1\right)^2}{a+b+c+d}=\dfrac{16}{a+b+c+d}\)
Dấu = xảy rakhi a=b=c=d
CM : bn tự chứng minh
Áp dụng:
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{4}{c}+\dfrac{16}{d}=\dfrac{1^2}{a}+\dfrac{1^2}{b}+\dfrac{2^2}{c}+\dfrac{4^2}{d}\ge\dfrac{\left(1+1+2+4\right)^2}{a+b+c+d}=\dfrac{64}{a+b+c+d}\)
Dấu = xảy ra khi \(a=b=\dfrac{c}{2}=\dfrac{d}{4}\)
Sửa: CMR \(\dfrac{a^3+c^3+m^3}{b^3+d^3+n^3}=\left(\dfrac{a+c-m}{b+d-n}\right)^3\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{m}{n}=k\Rightarrow a=kb;c=kd;m=kn\)
\(\dfrac{a^3+c^3+m^3}{b^3+d^3+n^3}=\dfrac{k^3b^3+k^3d^3+k^3n^3}{b^3+d^3+n^3}=\dfrac{k^3\left(b^3+d^3+n^3\right)}{b^3+d^3+n^3}=k^3\)
\(\left(\dfrac{a+c-m}{b+d-m}\right)^3=\left(\dfrac{kb+kd-kn}{b+d-n}\right)^3=\left(\dfrac{k\left(b+d-n\right)}{b+d-n}\right)^3=k^3\)
\(\Rightarrow\dfrac{a^3+c^3+m^3}{b^3+d^3+n^3}=\left(\dfrac{a+c-m}{b+d-n}\right)^3\left(=k^3\right)\)
\(VT=\dfrac{\left(a+c\right)^2}{\left(a+c\right)\left(a+b\right)}+\dfrac{\left(b+d\right)^2}{\left(b+c\right)\left(b+d\right)}+\dfrac{\left(c+a\right)^2}{\left(c+a\right)\left(c+d\right)}+\dfrac{\left(d+b\right)^2}{\left(d+a\right)\left(d+b\right)}\)
\(VT\ge\dfrac{\left(2a+2b+2c+2d\right)^2}{\left(a+b\right)\left(a+c\right)+\left(b+c\right)\left(b+d\right)+\left(a+c\right)\left(c+d\right)+\left(a+d\right)\left(b+d\right)}=\dfrac{4\left(a+b+c+d\right)^2}{\left(a+b+c+d\right)^2}=4\)
Dấu "=" xảy ra khi \(a=b=c=d\)
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+d}+\dfrac{d}{d+a}=2\)
\(1-\dfrac{a}{a+b}-\dfrac{b}{b+c}+1-\dfrac{c}{c+d}-\dfrac{d}{d+a}=0\)
\(\dfrac{b}{a+b}-\dfrac{b}{b+c}+\dfrac{d}{c+d}-\dfrac{d}{d+a}=0\)
\(\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)
<=>b(c+d)(d+a)+d(a+b)(b+c)=0 (vì c≠a)
<=>abc-acd+bd2-b2d=0
<=> (b-d)(ac-bd)=0 <=> ac - bd =0 (vì b≠d) <=> ac = bd
Vậy abcd =(ac)(bd)=(ac)2
Lời giải:
Ta có:
\(M=\frac{a}{a+b+c}+\frac{b}{a+b+d}+\frac{c}{b+c+d}+\frac{d}{a+d+c}\)
\(> \frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}\)
\(\Leftrightarrow M>\frac{a+b+c+d}{a+b+c+d}=1(1)\)
Mặt khác:
\(M=1-\frac{b+c}{a+b+c}+1-\frac{a+d}{a+b+d}+1-\frac{b+d}{b+c+d}+1-\frac{a+c}{a+d+c}\)
\(\Leftrightarrow M=4-\underbrace{\left(\frac{b+c}{a+b+c}+\frac{a+d}{a+b+d}+\frac{b+d}{b+c+d}+\frac{a+c}{a+d+c}\right)}_{N}\)
Có: \(N>\frac{b+c}{a+b+c+d}+\frac{a+d}{a+b+c+d}+\frac{b+d}{a+b+c+d}+\frac{a+c}{a+b+c+d}\)
\(\Leftrightarrow N>\frac{2(a+b+c+d)}{a+b+c+d}=2\)
\(\Rightarrow M=4-N< 4-2\Leftrightarrow M< 2(2)\)
Từ \((1);(2)\Rightarrow 1< M< 2\Rightarrow M\not\in \mathbb{N}\)