Question

bài 1chứng minh bất đẳng thức

a,a^2 + b^2 \(\ge\)1/2 vs a+b=1

b,a^2+b^2+c^2\(\ge\)1/3 vs a+b+c=1

bài 2a, cho a>0 ,b>a và 2a+b=0 tìm min của \(\dfrac{1}{2a}+\dfrac{1}{b}=1\)

b, cho a+b+c=4 tìm mã của ab+bc+ac

Answer 1

1a)\(a^2+b^2\ge\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{a^2+b^2}{2}\ge\dfrac{1}{4}\)(1)

Lại có:\(\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{4}=\dfrac{1}{4}\)

\(\Rightarrow\left(1\right)\) đúng\(\Rightarrowđpcm\)

1b)\(a^2+b^2+c^2\ge\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{a^2}{2}+\dfrac{b^2}{2}+\dfrac{c^2}{2}\ge\dfrac{1}{6}\)(2)

Lại có:\(\dfrac{a^2}{2}+\dfrac{b^2}{2}+\dfrac{c^2}{2}\ge\dfrac{\left(a+b+c\right)^2}{6}=\dfrac{1}{6}\)

\(\Rightarrow\left(2\right)\) đúng\(\Rightarrowđpcm\)

Answer 2

2b)Ta có:\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)(bđt phụ)

\(\Leftrightarrow ab+bc+ca\le\dfrac{4^2}{3}=\dfrac{16}{3}\)

\(\Rightarrow MAXA=\dfrac{16}{3}\Leftrightarrow x=y=z=\dfrac{4}{3}\)