Cho a, b, c > 0 thỏa mãn \(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}=2\). Tìm GTLN của N = abc
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12 tháng 4 2021
\(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}=2\)
=> \(\dfrac{1}{a+1}=1-\dfrac{1}{b+1}+1-\dfrac{1}{c+1}=\dfrac{b}{b+1}+\dfrac{c}{c+1}\ge2\sqrt{\dfrac{bc}{\left(b+1\right)\left(c+1\right)}}\)( AM-GM)
Tương tự ta có \(\dfrac{1}{b+1}\ge2\sqrt{\dfrac{ac}{\left(a+1\right)\left(c+1\right)}}\); \(\dfrac{1}{c+1}\ge2\sqrt{\dfrac{ab}{\left(a+1\right)\left(b+1\right)}}\)
Nhân vế với vế các bđt trên
=> \(\dfrac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge8\sqrt{\dfrac{a^2b^2c^2}{\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2}}=8\cdot\dfrac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\)
=> \(1\le8abc\)<=> \(abc\le\dfrac{1}{8}\)
Đẳng thức xảy ra <=> a=b=c=1/2
XO
30 tháng 12 2021
\(4M=\dfrac{4}{\left(a+b\right)+\left(a+c\right)}+\dfrac{4}{\left(a+b\right)+\left(b+c\right)}+\dfrac{4}{\left(c+a\right)+\left(b+c\right)}\)
\(\le\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{b+c}\)
\(=\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)
=> 8M \(\le\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)
\(\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}=2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=8\)
=> \(M\le1\)
Dấu "=" xảy ra <=> a = b = c = 3/4
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 12 2021
\(\dfrac{1}{2a+b+c}=\dfrac{1}{a+a+b+c}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{16}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Tương tự:
\(\dfrac{1}{a+2b+c}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{a+b+2c}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\right)\)
Cộng vế:
\(M\le\dfrac{1}{16}\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\)
\(M_{max}=1\) khi \(a=b=c=\dfrac{3}{4}\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 3 2023
\(Q=\dfrac{2a}{\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{\sqrt{c^2+ab+bc+ca}}\)
\(=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(=\sqrt{\dfrac{2a}{a+b}.\dfrac{2a}{a+c}}+\sqrt{\dfrac{2b}{a+b}.\dfrac{b}{2\left(b+c\right)}}+\sqrt{\dfrac{2c}{a+c}.\dfrac{c}{2\left(b+c\right)}}\)
\(\le\dfrac{1}{2}\left(\dfrac{2a}{a+b}+\dfrac{2a}{a+c}+\dfrac{2b}{a+b}+\dfrac{b}{2\left(b+c\right)}+\dfrac{2c}{a+c}+\dfrac{c}{2\left(b+c\right)}\right)\)
\(=\dfrac{9}{4}\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{7}{\sqrt{15}};\dfrac{1}{\sqrt{15}};\dfrac{1}{\sqrt{15}}\right)\)
3 tháng 4 2022
Bài 3:
\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)
\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 12 2021
\(a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2\ge2ab+2b+2=2\left(ab+b+1\right)\)
Tương tự ...
\(\Rightarrow P\le\dfrac{1}{2\left(ab+b+1\right)}+\dfrac{1}{2\left(bc+c+1\right)}+\dfrac{1}{2\left(ca+a+1\right)}\)
\(=\dfrac{1}{2}\left(\dfrac{c}{abc+bc+c}+\dfrac{1}{bc+c+1}+\dfrac{bc}{ca.bc+a.bc+bc}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{c}{1+bc+c}+\dfrac{1}{bc+c+1}+\dfrac{bc}{c+1+bc}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{c+1+bc}{1+bc+c}\right)=\dfrac{1}{2}\)
\(P_{max}=\dfrac{1}{2}\) khi \(a=b=c=1\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 1 2021
\(a^2-ab+b^2=\dfrac{1}{4}\left(a+b\right)^2+\dfrac{3}{4}\left(a-b\right)^2\ge\dfrac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow P\le\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
MH
5 tháng 5 2022
Áp dụng bđt Svácxơ, ta có:
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
\(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Áp dụng, thay vào A, ta có:
\(A\le\text{Σ}\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{3}{2}\)
Dấu "="⇔\(a=b=c=1\)
XO
11 tháng 2 2022
Ta có \(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}=2\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{\sqrt{ab}}=4\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=4-\dfrac{2}{\sqrt{ab}}\)
Khi đó P = \(\dfrac{1}{\sqrt{ab}}\left(4-\dfrac{2}{\sqrt{ab}}\right)=-2\left(\dfrac{1}{\sqrt{ab}}-1\right)^2+2\le2\)
Dấu "=" khi a = b = 1
22 tháng 12 2021
Ta có \(\sqrt{bc\left(1+a^2\right)}=\sqrt{bc+a^2bc}=\sqrt{bc+a\left(a+b+c\right)}\)
\(=\sqrt{\left(a+b\right)\left(a+c\right)}\)
Đặt BT đề cho là P
\(\Leftrightarrow P=\sum\dfrac{a}{\sqrt{bc\left(1+a^2\right)}}=\sum\sqrt{\dfrac{a}{a+b}\cdot\dfrac{a}{a+c}}\\ \Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{b+a}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)\\ \Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\cdot3=\dfrac{3}{2}\)
Dấu \("="\Leftrightarrow a=b=c=\sqrt{3}\)
\(\dfrac{1}{1+a}=1-\dfrac{1}{1+b}+1-\dfrac{1}{1+c}=\dfrac{b}{1+b}+\dfrac{c}{1+c}\ge2\sqrt{\dfrac{bc}{\left(1+b\right)\left(1+c\right)}}\)
Tương tự:
\(\dfrac{1}{1+b}\ge2\sqrt{\dfrac{ac}{\left(1+a\right)\left(1+c\right)}}\) ; \(\dfrac{1}{1+c}\ge2\sqrt{\dfrac{ab}{\left(1+a\right)\left(1+c\right)}}\)
Nhân vế với vế:
\(\dfrac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge\dfrac{8abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)
\(\Rightarrow abc\le\dfrac{1}{8}\)
\(N_{max}=\dfrac{1}{8}\) khi \(a=b=c=\dfrac{1}{2}\)