`(a^2+b^2)/2 >= ab`
`<=>a^2+b^2>=2ab`
`<=>a^2-2ab+b^2>=0`
`<=>(a-b)^2>=0, forall a,b`
\(\left(a-b\right)^2\ge0\)
\(\Leftrightarrow a^2+b^2-2ab\ge0\)
\(\Leftrightarrow\dfrac{a^2+b^2}{2}\ge ab\)
`(a^2+b^2)/2 >= ab`
`<=>a^2+b^2>=2ab`
`<=>a^2-2ab+b^2>=0`
`<=>(a-b)^2>=0, forall a,b`
\(\left(a-b\right)^2\ge0\)
\(\Leftrightarrow a^2+b^2-2ab\ge0\)
\(\Leftrightarrow\dfrac{a^2+b^2}{2}\ge ab\)
CM BĐT: \(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge\dfrac{2}{ab+1}\) với \(ab\ge1\)
Cho a và b không đồng thời bằng 0
Chứng minh \(\dfrac{a^2-ab+b^2}{a^2+ab+b^2}\ge\) \(\dfrac{1}{3}\)
CMR: \(\dfrac{1}{\left(1+a\right)^2}+\dfrac{1}{\left(1+b\right)^2}\ge\dfrac{1}{1+ab}\forall a,b\ge0\)
usechatgpt init success
BIết a + b + c = 3 CMR: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{3}{2}\)
Biết \(a+b=1\). Chứng minh rằng:
\(a,a^2+b^2\ge\dfrac{1}{2}\)
\(b,a^4+b^4\ge\dfrac{1}{8}\)
\(c,a^8+b^8\ge\dfrac{1}{128}\)
CMR:
a,\(\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{2}\)
b,Cho a+b=1,a>0,b>0 CMR:\(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\)\(\ge9\)
Cho 3 số dương a,b,c thỏa mãn
\(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\sqrt{2011}\)
CMR:\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)
Chứng minh rằng: \(\left(a^2+b^2+c^2\right)\left[\left(\dfrac{1}{\left(a-b\right)^2}+\dfrac{1}{\left(b-c\right)^2}+\dfrac{1}{\left(c-a\right)^2}\right)\right]\ge\dfrac{9}{2}\)