Cho
\(\frac{ab+1}{b}=\frac{bc+1}{c}=\frac{ac+1}{a}\)
C/m a=b=c
K
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31 tháng 10 2020
\(M=\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-a\right)}\)
Đánh giá đại diện: \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{\left(a-c\right)-\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}=\frac{1}{a-b}-\frac{1}{a-c}\)
Tương tự: \(\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{b-c}-\frac{1}{b-a}\)
\(\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{c-a}-\frac{1}{c-b}\)
\(\Rightarrow M=\frac{1}{a-b}-\frac{1}{a-c}+\frac{1}{b-c}-\frac{1}{b-a}+\frac{1}{c-a}-\frac{1}{c-b}\)
\(\Rightarrow M=\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}+\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}\)
\(\Rightarrow M=2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)=2N\left(đpcm\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 11 2019
\(VT=\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{1}{4}\left(\frac{ab}{a}+\frac{ab}{b}+\frac{bc}{b}+\frac{bc}{c}+\frac{ca}{c}+\frac{ca}{a}\right)\)
\(VT\le\frac{1}{4}\left(2a+2b+2c\right)=\frac{1}{2}\) (1)
Mặt khác \(\frac{bc}{a}+\frac{ac}{b}\ge2c\) ; \(\frac{bc}{a}+\frac{ab}{c}\ge2b\) ; \(\frac{ac}{b}+\frac{ab}{c}\ge2a\)
\(\Rightarrow2\left(\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\right)\ge2\left(a+b+c\right)\Rightarrow\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\ge1\)
\(\Rightarrow VP=\frac{1}{4}\left(1+\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\right)\ge\frac{1}{4}\left(1+1\right)=\frac{1}{2}\) (2)
Từ (1) và (2) suy ra đpcm
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 3 2020
\(P=\sum\frac{1}{\sqrt{a^2+b^2-ab+b^2+b^2+1}}\le\sum\frac{1}{\sqrt{ab+b^2+2b}}=\sum\frac{2}{\sqrt{4b\left(a+b+2\right)}}\)
\(\Rightarrow P\le\sum\left(\frac{1}{4b}+\frac{1}{a+b+1+1}\right)\le\sum\left(\frac{1}{4b}+\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+1+1\right)\right)\)
\(\Rightarrow P\le\frac{3}{8}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{3}{8}\le\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
2.
\(1\ge\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge\frac{9}{3+a+b+c}\)
\(\Rightarrow a+b+c+3\ge6\Rightarrow a+b+c\ge6\)
\(P=\sum\frac{a^3}{a^2+ab+b^2}=\sum\left(a-\frac{ab\left(a+b\right)}{a^2+ab+b^2}\right)\ge\sum\left(a-\frac{ab\left(a+b\right)}{3ab}\right)\)
\(\Rightarrow P\ge\sum\left(\frac{2a}{3}-\frac{b}{3}\right)=\frac{1}{3}\left(a+b+c\right)\ge\frac{6}{3}=2\)
Dấu "=" xảy ra khi \(a=b=c=2\)
10 tháng 3 2020
Ta có : \(ab\le\frac{a^2+b^2}{2}\)
\(\Rightarrow a^2-ab+3b^2+1\ge\frac{a^2}{2}+\frac{5}{2}b^2+1\)
Lại có : \(\left(\frac{a^2}{2}+\frac{5}{2}b^2+1\right)\left(\frac{1}{2}+\frac{5}{2}b^2+1\right)\ge\left(\frac{a}{2}+\frac{5}{2}b+1\right)^2\)
\(\Rightarrow\sqrt{a^2-ab+3b^2+1}\ge\frac{a}{4}+\frac{5b}{4}+\frac{1}{2}\)
\(\Rightarrow\frac{1}{\sqrt{a^2-ab+3b^2+1}}\le\frac{4}{a+b+b+b+b+b+1+1}\le\frac{4}{64}\left(\frac{1}{a}+\frac{5}{b}+2\right)\)
Khi đó :
\(P\le\frac{1}{16}\left(6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+6\right)\le\frac{3}{2}\)
Dấu " = " xay ra khi a=b=c=1
Vậy \(P_{Max}=\frac{3}{2}\) khi a=b=c=1