e)\(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(=1+\frac{b}{a}+\frac{a}{b}+1\)
\(=\left(1+1\right)+\left(\frac{a}{b}+\frac{b}{a}\right)\)
\(=2+\left(\frac{a.a}{b.a}+\frac{b.b}{a.b}\right)\)
\(=2+\frac{a.a+b.b}{b.a}\)
Vì \(\frac{a.a+b.b}{a.b}>=2\)
Nên \(2+\frac{a.a+b.b}{a.b}>=2+2=4\)
Hay \(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)>=4\)
a) \(a^2+b^2-2ab\)
\(=\left(a-b\right)^2\)
Vì \(\left(a-b\right)^2\) là binh phương của một số nên \(\left(a-b\right)^2>=0\)
Hay \(a^2+b^2-2ab>=0\)
c) \(a\left(a+2\right)\)
\(=a^2+2a\)
\(\left(a+1\right)^2\)
\(=\left(a+1\right)\left(a+1\right)\)
\(=a^2+a+a+1\)
\(=a^2+2a+1\)
Vì\(a^2+2a< a^2+2a+1\)
Nên \(a\left(a+2\right)< \left(a+1\right)^2\)
b, Áp dụng BDT cauchy , ta được
\(a^2+b^2\ge2.\sqrt{a^2b^2}\)
\(\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow\frac{a^2+b^2}{2}\ge ab\left(dpcm\right)\)
b) có: (a-b)2 >= 0
<=> a2-2ab+b2 >= 0
<=> a2+b2 >= 2ab
<=> (a2+b2)/2 >= ab
Hơi khó đọc nhưng thấy đúng thì tk nha
a) \(a^2+b^2-2ab=\left(a-b\right)^2\ge0\) ( luôn đúng )
Dấu " = " xảy ra <=> a=b
b) \(\frac{a^2+b^2}{2}\ge ab\)
\(\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow a^2+b^2-2ab\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)( luôn đúng )
Dấu " = " xảy ra <=> a=b
c) \(a\left(a+2\right)< \left(a+1\right)^2\)
\(\Leftrightarrow a^2+2a< a^2+2a+1\)
\(\Leftrightarrow0< 1\)( luôn đúng )
\(\Rightarrow a\left(a+2\right)< \left(a+1\right)^2\)
d) \(m^2+n^2+2\ge2\left(m+n\right)\)
\(\Leftrightarrow m^2+n^2+2-2n-2m\ge0\)
\(\Leftrightarrow\left(m-1\right)^2+\left(n-1\right)^2\ge0\)( luôn đúng )
Dấu " = " xảy ra <=> m=n=1
e) Áp dụng BĐT AM-GM ta có:
\(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge2.\sqrt{ab}.\frac{2}{\sqrt{ab}}=4\)
Dấu " = " xảy ra <=> a=b
