cho a,b,c là các số dương thỏa mãn abc=1
tìm GTLN của M= 1/a^2+2b^2+3 + 1/b^2+2c^2+3 + 1/c^2+2a^2+3
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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 \(a=b=c=1\)
L
1
26 tháng 4 2020
BĐT cần chứng minh tương đương với :
\(\left(a^2b+b^2c+c^2a\right)\left(2+\frac{1}{a^2b^2c^2}\right)\ge9\)
\(\Leftrightarrow2\left(a^2b+b^2c+c^2a\right)+\frac{1}{ab^2}+\frac{1}{bc^2}+\frac{1}{ca^2}\ge9\)
Áp dụng BĐT Cô-si cho 3 số dương ,ta có :
\(a^2b+a^2b+\frac{1}{ab^2}\ge3\sqrt[3]{a^2b.a^2b.\frac{1}{ab^2}}=3a\)
tương tự : \(b^2c+bc^2+\frac{1}{bc^2}\ge3b\), \(\left(c^2a+ca^2+\frac{1}{ca^2}\right)\ge3c\)
Cộng 3 BĐT trên theo vế, ta được :
\(2\left(a^2b+b^2c+c^2a\right)+\frac{1}{ab^2}+\frac{1}{bc^2}+\frac{1}{ca^2}\ge3\left(a+b+c\right)=9\)
Dấu "=" xảy ra khi a = b = c = 1
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 8 2021
\(\dfrac{1}{\left(a+b+a+c\right)^2}\le\dfrac{1}{4\left(a+b\right)\left(a+c\right)}=\dfrac{1}{4\left(a^2+ab+bc+ca\right)}\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=\dfrac{1}{64}\left(\dfrac{2}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)
Tương tự và cộng lại:
\(P\le\dfrac{1}{64}\left(\dfrac{4}{a^2}+\dfrac{4}{b^2}+\dfrac{4}{c^2}\right)=\dfrac{1}{16}.3=\dfrac{3}{16}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
24 tháng 8 2021
Áp dụng bđt: \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(1\right)\)
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\)
\(\Rightarrow P\le\dfrac{1}{16}\left[\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)^2+\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)^2+\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)^2\right]\)\(\Rightarrow16P\le\dfrac{2}{\left(a+b\right)^2}+\dfrac{2}{\left(b+c\right)^2}+\dfrac{2}{\left(a+c\right)^2}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(b+c\right)\left(c+a\right)}\)
Áp dụng: \(x^2+y^2+z^2\ge xy+yz+xz\left(2\right)\) với a+b=x,b+c=y,c+a=z
\(\Rightarrow16P\le\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}\)
Ta có: \(\dfrac{1}{\left(a+b\right)^2}\le4.16.\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)(do (1))
\(\Rightarrow16P\le\dfrac{1}{4}.16\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2+\left(\dfrac{1}{b}+\dfrac{1}{c}\right)^2+\left(\dfrac{1}{c}+\dfrac{1}{a}\right)^2\right]=\dfrac{1}{4}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\right)\le\dfrac{1}{4}.4.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=3\)(do(2) và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=3\))
\(\Rightarrow P\le\dfrac{3}{16}\)
\(ĐTXR\Leftrightarrow a=b=c=1\)
7 tháng 5 2021
Ta có:
\(\frac{1}{a^2+2b^2+3}=\frac{1}{\left(a^2+b^2\right)+\left(b^2+1\right)+2}\le\frac{1}{2ab+2b+2}=\frac{1}{2}\cdot\frac{1}{ab+b+1}\)
Tương tự CM được:
\(\frac{1}{b^2+2c^2+3}\le\frac{1}{2}\cdot\frac{1}{bc+c+1}\) và \(\frac{1}{c^2+2a^2+3}\le\frac{1}{2}\cdot\frac{1}{ca+a+1}\)
\(\Rightarrow VT\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{ab^2c+abc+ab}+\frac{b}{abc+ab+b}\right)\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{b+1+ab}+\frac{b}{1+ab+b}\right)=\frac{1}{2}\cdot1=\frac{1}{2}\)
Dấu "=" xảy ra khi: a = b = c = 1
7 tháng 5 2021
A=\(\frac{1}{a^2+2b^2+3}\)+\(\frac{1}{b^2+2c^2+3}\)+\(\frac{1}{c^2+2a^2+3}\)
ta có: \(\frac{1}{a^2+2b^2+3}\)=\(\frac{1}{\left(a^2+b^2\right)+\left(b^2+1\right)+2}\)\(\le\)\(\frac{1}{2\left(ab+b+1\right)}\)
vì : a2+b2\(\ge\)2\(\sqrt{a^2b^2}\)=2ab
b2+1\(\ge\)2\(\sqrt{b^2x1}\)=2b
cmtt => A\(\le\)\(\frac{1}{2}\)x(\(\frac{1}{ab+b+1}\)+\(\frac{1}{bc+c+1}\)+\(\frac{1}{ca+a+1}\))
=\(\frac{1}{2}\)x(\(\frac{1}{ab+b+1}\)+\(\frac{ab}{ab^2c+abc+ab}\)+\(\frac{b}{cba+ab+b}\))
=\(\frac{1}{2}\)x (\(\frac{1}{ab+b+1}\)+\(\frac{ab}{ab+b+1}\)+\(\frac{b}{ab+b+1}\))=\(\frac{1}{2}\)x\(\frac{ab+b+1}{ab+b+1}\)=\(\frac{1}{2}\)
dấu "=" xảy ra <=> a=b=c=1
21 tháng 3 2021
Ta có: \(\dfrac{a^3}{a^2+2b^2}=a-\dfrac{2ab^2}{a^2+2b^2}\ge a-\dfrac{2ab^2}{3\sqrt[3]{a^2b^4}}=a-\dfrac{2}{3}\sqrt[3]{ab^2}\ge a-\dfrac{2}{9}\left(a+b+b\right)=a-\dfrac{2}{9}\left(a+2b\right)\) Chứng minh tương tự ta được:
\(\dfrac{b^3}{b^2+2c^2}\ge b-\dfrac{2}{9}\left(b+2c\right);\dfrac{c^3}{c^2+2a^2}\ge c-\dfrac{2}{9}\left(c+2a\right)\)
\(\Rightarrow\dfrac{a^3}{a^2+2b^2}+\dfrac{b^3}{b^2+2c^2}+\dfrac{c^3}{c^2+2a^2}\ge a+b+c-\dfrac{2}{9}\left(a+2b+b+2c+c+2a\right)=a+b+c-\dfrac{2}{9}\left(3a+3b+3c\right)=\dfrac{1}{3}\left(a+b+c\right)\ge\dfrac{1}{3}\cdot3\sqrt[3]{abc}=1\)Dấu = xảy ra \(\Leftrightarrow a=b=c=1\)
28 tháng 4 2020
Đặt \(a=x^2;b=y^2;c=z^2\)khi đó ta được xyz=1 và biểu thức P viết được thành
\(P=\frac{1}{x^2+2y^2+3}+\frac{1}{y^2+2x^2+3}+\frac{1}{z^2+2x^2+3}\)
Ta có \(x^2+y^2\ge2xy;y^2+1\ge2y\Rightarrow x^2+2y^2+3\ge2\left(xy+y+1\right)\)
Do đó ta được \(\frac{1}{x^2+2y^2+3}\le\frac{1}{2}\cdot\frac{1}{xy+y+1}\)
Chứng minh tương tự ta có:
\(\frac{1}{y^2+2z^2+3}\le\frac{1}{2}\cdot\frac{1}{yz+z+1};\frac{1}{z^2+2x^2+3}\le\frac{1}{2}\cdot\frac{1}{zx+z+1}\)
Cộng các vế BĐT trên ta được
\(P\le\frac{1}{2}\left(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}\right)\)
Ta cần chứng minh \(\frac{1}{ab+b+1}+\frac{1}{bc+b+1}+\frac{1}{ca+a+1}=1\)
Do xyz=1 nên ta được
\(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}=\frac{zx}{z+1+zx}+\frac{x}{1+zx+z}+\frac{1}{zx+x+1}=1\)
Từ đó ta được
\(P\le\frac{1}{2}\). Dấu "=" xảy ra <=> a=b=c=1
Áp dụng bđt cô si ta có:
\(a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2\ge2ab+2b+2=2\left(ab+b+1\right)\)
\(b^2+2c^2+3\ge2\left(bc+c+1\right)\)
\(c^2+2a^2+3\ge2\left(ac+a+1\right)\)
=> \(M\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{bcab+abc+ab}+\frac{b}{abc+ab+b}\right)\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{b+1+ab}+\frac{b}{1+ab+b}\right)\)
\(=\frac{1}{2}.\frac{ab+b+1}{ab+b+1}=\frac{1}{2}\)
Bổ sung:
Dấu "=" xảy ra <=> a = b = c = 1
Vậy GTLN của M = 1/2 tại a = b = c = 1.