Question

Chứng minh bất đẳng thức : ( a + b )² ≤ 2( a² + b²)

Answer 1

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

\(\Leftrightarrow2a^2+2b^2\ge a^2+b^2+2ab\)

\(\Leftrightarrow2a^2+2b^2-a^2-b^2-2ab\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)

=>Đpcm

Answer 2

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

\(\Leftrightarrow a^2+2ab+b^2\le2a^2+2b^2\)

\(\Leftrightarrow a^2+2ab+b^2-2a^2-2b^2\le0\)

\(\Leftrightarrow-\left(a^2-2ab+b^2\right)\le0\)

\(\Leftrightarrow-\left(a-b\right)^2\le0\) ( luôn đúng )

dấu " = " xảy ra khi a = b

Answer 3

xoắn:v Bunyakovsky: \(\left(a+b\right)^2\le\left(1^2+1^2\right)\left(a^2+b^2\right)=2\left(a^2+b^2\right)\)

Answer 4

\(\left(a+b\right)^2\le2\left(a^2+b^2\right)\left(1\right)\\ \Leftrightarrow a^2+2ab+b^2-2a^2-2b^2\le0\\ \Leftrightarrow-a^2+2ab-b^2\le0\\ \Leftrightarrow-\left(a^2-2ab+b^2\right)\le0\\ \Leftrightarrow-\left(a-b\right)^2\le0\left(2\right)\)

Bất đẳng thức (2) luôn đúng \(\forall a;b\)

nên bất đẳng thức (1) cũng luôn đúng \(\forall a;b\)

Vậy \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\)

đẳng thức xảy ra khi: \(a=b\)