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Từ gt⇒0≤b≤2−2a3≤2;0≤b≤4−2a≤4⇒0≤b≤2−2a3≤2;0≤b≤4−2a≤4
⇒0≤b≤2⇒0≤b≤2
Tương tự⇒a,b∈[0;2]⇒a,b∈[0;2]
Ta có:
A=a(a−2)−b≤a(a−2)≤0A=a(a−2)−b≤a(a−2)≤0
Dấu = xảy ra⇔a=b=0⇔a=b=0 hoặc a=2,b=0a=2,b=0
Ta có:
A≥a2−2a+2a3−2=(a−23)2−229≥−229A≥a2−2a+2a3−2=(a−23)2−229≥−229
và A≥a2−2a+2a−4=a2−4≥−4A≥a2−2a+2a−4=a2−4≥−4
Vì A≥−4A≥−4 ko xảy ra dấu = nên A≥−229⇔a=23,b=149
Ta có :\(\left(a-b\right)^2\ge0\forall a;b\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow2a^2+2b^2\ge a^2+b^2+2ab\Leftrightarrow2a^2+2b^2\ge\left(a+b\right)^2\)
Suy ra \(\frac{2011}{2a^2+2b^2+2008}\le\frac{2011}{\left(a+b\right)^2+2008}=\frac{2011}{4+2008}=\frac{2011}{2012}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=1\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{bc}{a+3b+2c}\le\frac{1}{9}\left(\frac{bc}{a+b}+\frac{bc}{b+c}+\frac{c}{2}\right)\)
\(\frac{ca}{b+3c+2a}\le\frac{1}{9}\left(\frac{ca}{b+c}+\frac{ca}{c+a}+\frac{a}{2}\right)\)
\(\frac{ab}{c+3a+2b}\le\frac{1}{9}\left(\frac{ab}{c+a}+\frac{ab}{a+b}+\frac{b}{2}\right)\)
Cộng theo vế của 3 BĐT ta có:
\(VT\le\frac{1}{9}\left(\frac{a+b+c}{2}+\frac{ca+ab}{a+c}+\frac{ab+bc}{a+b}+\frac{bc+ca}{b+c}\right)\)
\(=\frac{1}{9}\left(a+b+c+\frac{a+b+c}{2}\right)=1\)
Dấu "=" khi a=b=c=2
a) ta có : \(2A+3B=0\) \(\Leftrightarrow2.\dfrac{5}{2m+1}+3.\dfrac{4}{2m-1}=0\)
\(\Leftrightarrow\dfrac{10}{2m+1}+\dfrac{12}{2m-1}=0\Leftrightarrow\dfrac{10\left(2m-1\right)+12\left(2m+1\right)}{\left(2m+1\right)\left(2m-1\right)}=0\)
\(\Leftrightarrow\dfrac{20m-10+24m+12}{4m^2-1}=0\Leftrightarrow\dfrac{44m+2}{4m^2-1}=0\)
\(\Leftrightarrow44m+2=0\Leftrightarrow44m=-2\Leftrightarrow m=\dfrac{-2}{44}=\dfrac{-1}{22}\) vậy \(m=\dfrac{-1}{22}\)
b) ta có : \(AB=\dfrac{5}{2m+1}.\dfrac{4}{2m-1}=\dfrac{5.4}{\left(2m+1\right)\left(2m-1\right)}\)
ta có : \(A+B=\dfrac{5}{2m+1}+\dfrac{4}{2m-1}=\dfrac{5\left(2m-1\right)+4\left(2m+1\right)}{\left(2m+1\right)\left(2m-1\right)}\)
\(\Rightarrow AB=A+B\Leftrightarrow\dfrac{5.4}{\left(2m+1\right)\left(2m-1\right)}=\dfrac{5\left(2m-1\right)+4\left(2m+1\right)}{\left(2m+1\right)\left(2m-1\right)}\)
\(\Leftrightarrow5.4=5\left(2m-1\right)+4\left(2m+1\right)\Leftrightarrow20=10m-5+8m+4\)
\(\Leftrightarrow20=18m-1\Leftrightarrow18m=20+1=21\Leftrightarrow m=\dfrac{21}{18}=\dfrac{7}{6}\) vậy \(m=\dfrac{7}{6}\)
