Question

Cho\(a^2+b^2=1\). Chứng minh\(\left(a+b\right)^2\le2\)

Answer 1

Ta có:\(\left(a+b\right)^2\)

\(=a^2+2ab+b^2\)

\(=1+2ab\)(do \(a^2+b^2=1\))

Lại có:\(1=a^2+b^2\ge2ab\)

\(\Leftrightarrow2ab\le1\)

\(\Leftrightarrow a^2+2ab+b^2\le1+1\)

\(\Leftrightarrow\left(a+b\right)^2\le2\left(đpcm\right)\)

Answer 2

Bo de : \(h^2+anh^2\ge\frac{\left(h+anh\right)^2}{2}\)

\(< =>\left(h-anh\right)^2\ge0\)(right)

Dau "=" xay ra \(< =>h=anh\)

Ap dung \(1=a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\)\(< =>\left(a+b\right)^2\le2\)

Dau "=" xay ra \(< =>a=b=\frac{1}{\sqrt{2}}\)

Answer 3

áp dụng bunhia - cốp xki với bộ số (a,b)(1,1)

\(\left(a^2+b^2\right)\left(1+1\right)>=\left(a+b\right)^2\)

\(2>=\left(a+b\right)^2\)

<=> đpcm

\(\)