a) Ta có: \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)
\(< =>2x^2+2y^2\ge x^2+2xy+y^2\)
\(< =>x^2+y^2\ge2xy\)
\(< =>x^2-2xy+y^2\ge0\)
\(< =>\left(x-y\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra <=> x=y
=>(đpcm).
a. \(x^2+y^2-\frac{\left(x+y\right)^2}{2}\ge0\)
\(\Leftrightarrow2x^2+2y^2-\left(x+y\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2-x^2-2xy-y^2\ge0\)
\(\Leftrightarrow x^2-2xy+y^2=\left(x+y\right)^2\ge0\) (Luôn đúng)
Hay \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\left(Dfcm\right)\)
b. \(ab-\frac{\left(a+b\right)^2}{4}\le0\)
\(\Leftrightarrow4ab-a^2-2ab-b^2\le0\)
\(\Leftrightarrow-\left(a^2-2ab+b^2\right)=-\left(a-b\right)^2\le0\) (Luôn đúng)
Hay \(ab\le\frac{\left(a+b\right)^2}{4}\)
a)\(< =>\frac{2x^2+2y^2-x^2-2xy-y^2}{2}\ge0\)<=>\(\frac{x^2-2xy+y^2}{2}\ge0< =>\frac{\left(x-y\right)^2}{2}\ge0\left(lđ\right)\)
b)<=>\(\frac{4ab-a^2-2ab-b^2}{4}\le0< =>\frac{-\left(a^2-2ab+b^2\right)}{4}\le0\)<=>\(\frac{-\left(a-b\right)^2}{4}\le0\left(lđ\right)\)
b) Ta có: \(ab\le\frac{\left(a+b\right)^2}{4}\)
\(< =>a^2+2ab+b^2\ge4ab\)
\(< =>a^2+2ab-4ab+b^2\ge0\)
\(< =>a^2-2ab+b^2\ge0\)
\(< =>\left(a-b\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra <=> a=b.
=> (đpcm).
Có chỗ nào ko hiểu thì cứ hỏi lại nha bn ^-^
