Question

Cho a,b,c>0 chứng minh rằng :

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

b) \(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\le\dfrac{a+b+c}{2}\)

Answer 1

Áp dụng BĐT Cauchy ta có

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

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

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

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

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

Dấu bằng xảy ra khi a=b=c

Làm tắt vài chỗ thông cảm

Answer 2

Câu b,

Ta có BĐT Cauchy \(a^2+b^2\ge2ab\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\)

\(\Rightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\)

\(\Rightarrow\dfrac{ab}{a+b}\le\dfrac{\left(a+b\right)^2}{4\left(a+b\right)}=\dfrac{a+b}{4}\)

Tương tự \(\dfrac{bc}{b+c}\le\dfrac{b+c}{4}\)

\(\dfrac{ac}{a+c}\le\dfrac{a+c}{4}\)

Cộng theo vế ta đc \(VT\le\dfrac{2\left(a+b+c\right)}{4}=\dfrac{a+b+c}{2}\)

Dấu bằng xảy ra khi a=b=c