Question

Chứng minh rằng:Nếu a,b,c > 0 thì: \(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\le\dfrac{a+b+c}{2}\)

Answer 1

Áp dụng BĐT BSC:

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

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

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

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

Đẳng thức xảy ra khi \(a=b=c\)

Answer 2

4ab ≤ (a + b)² ⇒ \(\dfrac{4ab}{a+b}\le a+b\)

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

⇒ Cộng lại vế với vế :

4VT ≤ 2 (a+b+c) ⇒ VT ≤ \(\dfrac{a+b+c}{2}\)