cho a, b, c thỏa mãn a+b+c=79, \(\frac{a+7}{a-7}=\frac{b+4}{b-4},\frac{b+5}{b-5}=\frac{c+6}{c-6}\)
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NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 6 2019
Ta có đánh giá: \(\frac{a^7+b^7}{a^5+b^5}\ge\frac{a^2+b^2}{2}\)
\(\Leftrightarrow2a^7+2b^7\ge a^7+b^7+a^5b^2+a^2b^5\)
\(\Leftrightarrow a^5\left(a^2-b^2\right)-b^5\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\left(a^4+a^3b+a^2b^2+ab^3+b^4\right)\ge0\) (luôn đúng)
Tương tự \(\frac{b^7+c^7}{b^5+c^5}\ge\frac{b^2+c^2}{2}\) ; \(\frac{c^7+a^7}{c^5+a^5}\ge\frac{a^2+c^2}{2}\)
\(\Rightarrow VT\ge a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}=\frac{1}{3}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
18 tháng 6 2019
Áp dụng bđt Cauchy-Schwarz :
\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\)
Đề thiếu không bạn ?
30 tháng 7 2019
Nhân 2 vế của 2 ĐT đề bài ta có
\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)=\frac{47}{10}\)
<=> \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\left(\frac{a}{a+b}+\frac{b}{a+b}\right)+\left(\frac{b}{b+c}+\frac{c}{b+c}\right)+\left(\frac{c}{a+c}+\frac{a}{a+c}\right)=\frac{47}{10}\)
=>\(P=\frac{17}{10}\)
Vậy \(P=\frac{17}{10}\)
TD
13 tháng 2 2017
Đặt \(\frac{a}{2009}=\frac{b}{2010}=\frac{c}{2011}=k\left(k\ne0\right)\)
\(\Rightarrow a=2009k;b=2010k;c=2011k\)
\(\Rightarrow M=4\left(2009k-2010k\right)\left(2010k-2011l\right)-\left(2011k-2009k\right)^2\)\(\Rightarrow M=4\left(-k\right)\left(-k\right)-\left(2k\right)^2\)
\(\Rightarrow M=4k^2-4k^2=0\)
Vậy M=0
21 tháng 11 2019
Đặt \(P=\frac{a^4}{\left(a+2\right)\left(b+2\right)}+\frac{b^4}{\left(b+2\right)\left(c+2\right)}+\frac{c^4}{\left(c+2\right)\left(a+2\right)}\)
Áp dụng BĐT AM-GM ta có:
\(\frac{a^4}{\left(a+2\right)\left(b+2\right)}+\frac{a+2}{27}+\frac{b+2}{27}+\frac{1}{9}\ge4\sqrt[4]{\frac{a^2}{\left(a+2\right)\left(b+2\right)}.\frac{a+2}{27}.\frac{b+2}{27}.\frac{1}{9}}=\frac{4a}{9}\)(1)
\(\frac{b^4}{\left(b+2\right)\left(c+2\right)}+\frac{b+2}{27}+\frac{c+2}{27}+\frac{1}{9}\ge4\sqrt[4]{\frac{b^2}{\left(b+2\right)\left(c+2\right)}.\frac{b+2}{27}.\frac{c+2}{27}.\frac{1}{9}}=\frac{4b}{9}\)(2)
\(\frac{c^4}{\left(c+2\right)\left(a+2\right)}+\frac{c+2}{27}+\frac{a+2}{27}+\frac{1}{9}\ge4\sqrt[4]{\frac{c^2}{\left(c+2\right)\left(a+2\right)}.\frac{c+2}{27}.\frac{a+2}{27}.\frac{1}{9}}=\frac{4c}{9}\)(3)
Lấy \(\left(1\right)+\left(2\right)+\left(3\right)\)ta được:
\(P+\frac{2\left(a+b+c\right)+12}{27}+\frac{3}{9}\ge\frac{4\left(a+b+c\right)}{9}\)
\(\Leftrightarrow P+\frac{2}{3}+\frac{3}{9}\ge\frac{4}{3}\)
\(\Leftrightarrow P\ge\frac{1}{3}\left(đpcm\right)\)Dấu"="xảy ra \(\Leftrightarrow a=b=c=1\)
19 tháng 3 2020
Ta có \(\left(a-1\right)^2\left(a^2+a+1\right)\ge0\)\(\Leftrightarrow\left(a^2-2a+1\right)\left(a^2+a+1\right)\ge0\)
\(\Leftrightarrow a^4-a^3-a+1\ge0\)
\(\Leftrightarrow a^4-a^3+1\ge a\)
\(\Leftrightarrow a^4-a^3+ab+2\ge a+ab+1\)
\(\Rightarrow\frac{1}{\sqrt{a^4-a^3+ab+2}}\le\frac{1}{\sqrt{ab+a+1}}\)
Tương tự \(\frac{1}{\sqrt{b^4-b^3+bc+2}}\le\frac{1}{\sqrt{bc+b+1}}\)
\(\frac{1}{\sqrt{c^4-c^3+ca+2}}\le\frac{1}{\sqrt{ca+c+1}}\)
Cộng từng vế các bđt trên ta được
\(VT\le\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\)
Áp dụng bđt Bunhiacopski ta có
\(VT\le\sqrt{3\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)}\)\(=\sqrt{3\left(\frac{1}{ab+a+1}+\frac{a}{abc+ab+a}+\frac{ab}{a^2bc+abc+ab}\right)}=\sqrt{3}\)
Dấu "=" xảy ra khi a=b=c=1
Ta có: \(\frac{a+7}{a-7}=\frac{b+4}{b-4}\Rightarrow\left(a+7\right)\left(b-4\right)=\left(a-7\right)\left(b+4\right)\)
\(\Rightarrow ab-4a+7b-28=ab+4a-7b-28\)
\(\Rightarrow-4a-4a+7b+7b=0\Rightarrow-8a+14b=0\)
\(\Rightarrow8a=14b\Rightarrow4a=7b\Rightarrow\frac{a}{7}=\frac{b}{4}\Rightarrow\frac{a}{35}=\frac{b}{20}\left(1\right)\)
Lại có: \(\frac{b+5}{b-5}=\frac{c+6}{c-6}\Rightarrow\left(b+5\right)\left(c-6\right)=\left(b-5\right)\left(c+6\right)\)
\(\Rightarrow bc-6b+5c-30=bc+6b-5c-30\)
\(\Rightarrow6b+6b=5c+5c\) => 12b = 10c
=>\(6b=5c\Rightarrow\frac{b}{5}=\frac{c}{6}\Rightarrow\frac{b}{20}=\frac{c}{24}\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow\frac{a}{35}=\frac{b}{20}=\frac{c}{24}=\frac{a+b+c}{35+20+24}=\frac{79}{79}=1\)
=>a=35,b=20,c=24
cảm ơn bạn