Question

Cho a+b+c=2p

CMR: (p-a)(p-b)(p-c)<\(\dfrac{1}{8}\)abc

Answer 1

Áp dụng bất đẳng thức Cô - si ta có:

\(\left(p-a\right)\left(p-b\right)\le\dfrac{(p-a+p-b)^2}{4}=\dfrac{\left(2p-a-b\right)^2}{4}=\dfrac{c^2}{4}\)

\(\left(p-a\right)\left(p-c\right)\le\dfrac{(p-a+p-c)^2}{4}=\dfrac{\left(2p-a-c\right)^2}{4}=\dfrac{b^2}{4}\)

\(\left(p-b\right)\left(p-c\right)\le\dfrac{(p-b+p-c)^2}{4}=\dfrac{\left(2p-b-c\right)^2}{4}=\dfrac{a^2}{4}\)

\(\Rightarrow\left[\left(p-a\right)\left(p-b\right)\left(p-c\right)\right]^2\le\dfrac{a^2b^2c^2}{64}\)

\(\Leftrightarrow\left(p-a\right)\left(p-b\right)\left(p-c\right)\le\dfrac{abc}{8}\) (đpcm)