b) Áp dụng bất đẳng thức Cosi ta có:
\(a^2+\dfrac{1}{4}\ge2\sqrt{\dfrac{a^2}{4}}=\left|a\right|\ge a\)
Tương tự : \(b^2+\dfrac{1}{4}\ge b\)
\(\Rightarrow a^2+b^2+\dfrac{1}{2}\ge a+b=1\Rightarrow a^2+b^2\ge\dfrac{1}{2}\)
Dấu = xảy ra khi a=b=1/2
a) Ta có: \(7x+4\ge5x-8\)
\(\Leftrightarrow7x-5x\ge-8-4\)
\(\Leftrightarrow2x\ge-12\)
\(\Leftrightarrow x\ge-6\)
Vậy: S={x|\(x\ge-6\)}
a)
\(7x+4\ge5x-8\\ 7x-5x\ge-8-4\\2x\ge-12\\ x\ge-6 \)
b,Áp dụng bất đẳng thức Bun-nhi-a-cốp-xki , ta có
\(\left(a^2+b^2\right)\left(1+1\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow\)2\(\left(a^2+b^2\right)\ge1\)
\(\Leftrightarrow\)\(\left(a^2+b^2\right)\ge\dfrac{1}{2}\)
