Question

Chứng minh \(\frac{x^2+y^2}{2}\ge xy\) với mọi x; y

Answer 1

Ta có:(x-y)²>0

=>x²-2xy+y²>0

=>x²+y²>2xy

=>x²+y²/2>xy

Dấu = xảy ra <=>x=y

Answer 2

\(\frac{x^2+y^2}{2}\ge xy\)

\(\Leftrightarrow\frac{x^2+y^2}{2}\ge\frac{2xy}{2}\)

\(\Leftrightarrow x^2+y^2\ge2xy\)

\(\Leftrightarrow x^2-2xy+y^2\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\left(\text{Luôn đúng với mọi x;y}\right)\)

\(\text{Vậy }\frac{x^2+y^2}{2}\ge xy\)

\(\text{Dấu "=" xảy ra khi }x=y\)

Answer 3

\(2.\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

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

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

\(\Leftrightarrow\left(a-b\right)^2\text{ (luôn đúng với mọi x;y)}\)

\(\text{Vậy }2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\text{và dấu "=" xảy ra khi }a=b\)