câu 1 :Cmr a)\(\frac{a^2+b^2}{2}\ge\left(\frac{a+b}{2}\right)^2\)
b) \(\frac{a^3+b^3+c^3}{3}\ge\left(\frac{a+b+c}{3}\right)^3\)
câu 2 : cho a+b=1 .Cm \(\frac{1}{a+1}+\frac{1}{b+1}\ge\frac{4}{3}\)
câu 3: cho a+b+c=1và a,b,c>0.CMR \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)
câu 4 Tim max của : ab+2(a+b) ...biết a2+b2=1
Câu 1: a)
b) Áp dụng Bđt Holder ta có:
\(\Rightarrow9\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)^3\)
\(\Rightarrow\frac{a^3+b^3+c^3}{3}\ge\frac{\left(a+b+c\right)^3}{27}=\left(\frac{a+b+c}{3}\right)^3\)(đpcm)
Dấu = khi a=b=c
Câu 2:
Áp dụng Bđt \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)ta có:
\(\frac{1}{a+1}+\frac{1}{b+1}\ge\frac{4}{a+b+1+1}=\frac{4}{3}\)(Đpcm)
Dấu = khi \(a=b=\frac{1}{2}\)
Câu 3:
Áp dụng Bđt \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=9\left(a+b+c=1\right)\)(Đpcm)
Dấu = khi \(a=b=c=\frac{1}{3}\)
Câu 4: nghĩ sau