vì a;b>0\(\Rightarrow a+b>=2\sqrt{ab}\Rightarrow1>=2\sqrt{ab}\Rightarrow\frac{1}{2}>=\sqrt{ab}\Rightarrow\frac{1}{4}>=ab\)(bđt cosi)
dấu = xảy ra khi a=b=\(\frac{1}{2}\)
\(M=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2=1+\frac{2}{a}+\frac{1}{a^2}+1+\frac{2}{b}+\frac{1}{b^2}\)
\(=2+\left(\frac{2}{a}+\frac{2}{b}\right)+\left(\frac{1}{a^2}+\frac{1}{b^2}\right)>=2+2\sqrt{\frac{2}{a}\cdot\frac{2}{b}}+2\cdot\sqrt{\frac{1}{a^2}\cdot\frac{1}{b^2}}\)(bđt cosi )
dấu = xảy ra khi \(\frac{2}{a}=\frac{2}{b}\Rightarrow a=b=\frac{1}{2};\frac{1}{a^2}=\frac{1}{b^2}\Rightarrow a=b=\frac{1}{2}\)\(\Rightarrow\)dấu = xảy ra khi \(a=b=\frac{1}{2}\)
\(=2+\frac{4}{\sqrt{ab}}+\frac{2}{\sqrt{a^2b^2}}=2+\frac{4}{\sqrt{ab}}+\frac{2}{ab}>=2+\frac{4}{\frac{1}{2}}+\frac{2}{\frac{1}{4}}=2+8+8=18\)
\(\Rightarrow M>=18\Rightarrow\)min M là 18
vậy min M là 18 khi a=b=\(\frac{1}{2}\)
Áp dụng bđt Cauchy-Schwarz dạng Engel ta có :
\(M=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2=\frac{\left(1+\frac{1}{a}\right)^2}{1}+\frac{\left(1+\frac{1}{b}\right)^2}{1}\ge\frac{\left(1+\frac{1}{a}+1+\frac{1}{b}\right)^2}{2}=\frac{\left(2+\frac{1}{a}+\frac{1}{b}\right)}{2}\)(1)
Lại có \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}=4\)(2)
Từ (1) và (2) => \(M=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2\ge\frac{\left(2+\frac{1}{a}+\frac{1}{b}\right)^2}{2}\ge\frac{\left(2+4\right)^2}{2}=18\)
Đẳng thức xảy ra khi a = b = 1/2
Vậy MinM = 18, đạt được khi a = b = 1/2
