Question

Cho a, b, c là ba cạnh của tam giác , p là nửa chu vi tam giác đó. Chứng minh rằng 

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

Answer 1

                                  Lời giải

Theo đề bài thì \(p=\frac{a+b+c}{2}\Rightarrow p-a=\frac{a+b+c}{2}-a=\frac{b+c-a}{2}\)

Tương tự: \(p-b=\frac{c+a-b}{2};p-c=\frac{a+b-c}{2}\)

Ta cần c/m: \(\frac{2}{b+c-a}+\frac{2}{c+a-b}+\frac{2}{a+b-c}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Ta có: \(VT=\left(\frac{1}{b+c-a}+\frac{1}{c+a-b}\right)+\left(\frac{1}{c+a-b}+\frac{1}{a+b-c}\right)+\left(\frac{1}{a+b-c}+\frac{1}{b+c-a}\right)\)

\(\ge\frac{4}{2c}+\frac{4}{2a}+\frac{4}{2b}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{\left(đpcm\right)}\)

Answer 2

Ta có:\(p-a=\frac{a+b+c}{2}-a=\frac{b+c-a}{2}\Leftrightarrow\frac{1}{p-a}=\frac{2}{b+c-a}\)

Tương tự ta có:

\(\frac{1}{p-b}=\frac{2}{a+c-b}\)

\(\frac{1}{p-c}=\frac{2}{a+b-c}\)

\(\Rightarrow\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}=2\left(\frac{1}{a+c-b}+\frac{1}{b+c-a}+\frac{1}{a+b-c}\right)\)

Áp dụng bất đẳng thức Cauchy-Schwarz dạng engel ta có:

\(\frac{1}{b+c-a}=\frac{\left(1+1-1\right)^2}{b+c-a}\ge\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\)

Tương tự,ta có:

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

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

Cộng vế theo vế ta được:

\(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{đpcm}\)