Question

[Nguyễn Trung Hiếu]

Trong tam giác ABC. chứng minh rằng:
a) \(\left(p-a\right)\left(p-b\right)\left(p-c\right)\le\frac{1}{8}abc\)
b) \(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
c) \(R\ge2r\)
 

Answer 1

[Thắng Nguyễn]

a)Ta có:\(\left(p-a\right)\left(p-b\right)\le\frac{2p-b-a}{2}=\frac{c^2}{4}\)

Tương tự ta có: \(\left(p-a\right)\left(p-c\right)\le\frac{b^2}{4};\left(p-b\right)\left(p-c\right)\le\frac{c^2}{4}\)

\(\Rightarrow\left[\left(p-a\right)\left(p-b\right)\left(p-c\right)\right]^2\le\left(\frac{abc}{8}\right)^2\)

\(\Rightarrow\left(p-a\right)\left(p-b\right)\left(p-c\right)\le\frac{abc}{8}\)

b)\(VT=\frac{2}{-a+b+c}+\frac{2}{a-b+c}+\frac{2}{a+b-c}\)

\(=\frac{1}{-a+b+c}+\frac{1}{a-b+c}+\frac{1}{a+b-c}+\frac{1}{-a+b+c}+\frac{1}{a-b+c}+\frac{1}{a+b-c}\)

\(\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

c giải sau ăn cơm đã