Question

Co tam giác có nửa chu vi p = \(\dfrac{a+b+c}{2}\) với a , b , c là độ dài 3 cạnh

Chứng minh \(\dfrac{1}{p-a}\) + \(\dfrac{1}{p-b}\) + \(\dfrac{1}{p-c}\) \(\ge\) 2(\(\dfrac{1}{a}\) + \(\dfrac{1}{b}\) + \(\dfrac{1}{c}\) )

Answer 1

Từ \(p=\dfrac{a+b+c}{2}\Rightarrow2p=a+b+c\)

Áp dụng BĐT \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{\left(1+1\right)^2}{a+b}=\dfrac{4}{a+b}\) ta có:

\(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{p-a+p-b}=\dfrac{4}{2q-a-b}\)

\(=\dfrac{4}{a+b+c-a-b}=\dfrac{4}{c}\). Tương tự cho 2 BĐT còn lại:

\(\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{a};\dfrac{1}{p-c}+\dfrac{1}{p-a}\ge\dfrac{4}{b}\)

Cộng theo vế 3 BĐT trên ta có:

\(2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Leftrightarrow\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Đẳng thức xảy ra khi \(a=b=c\)