Đáp án:
Giải thích các bước giải:
Đặt vế trái của BĐT đã cho là P
Ta có: $\dfrac{1}{a+b+2c}=\dfrac{1}{(a+c)+(b+c)} \leq \dfrac{1}{4}\left( \dfrac{1}{a+c}+\dfrac{1}{b+c}\right)$
$⇒\dfrac{ab}{a+b+2c} \leq ab\dfrac{1}{4}\left( \dfrac{1}{a+c}+\dfrac{1}{b+c}\right)=\dfrac{1}{4}\left( \dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)$
Hoàn toàn tương tự, ta có:
$\dfrac{bc}{b+c+2a}=\dfrac{bc}{a+b+a+c} \leq \dfrac{1}{4}\left( \dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right)$
$\dfrac{ca}{c+a+2b}=\dfrac{ca}{a+b+b+c} \leq \dfrac{1}{4}\left( \dfrac{ca}{a+b}+\dfrac{ca}{b+c}\right)$
Cộng vế với vế:
$⇒P \leq \dfrac{1}{4}\left( \dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)+\dfrac{1}{4}\left( \dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right)+\dfrac{1}{4}\left( \dfrac{ca}{a+b}+\dfrac{ca}{b+c}\right)$
$⇔P \leq \dfrac{1}{4}\left( \dfrac{ab}{a+c}+\dfrac{bc}{a+c}+\dfrac{ab}{b+c}+\dfrac{ca}{b+c}+\dfrac{bc}{a+b}+\dfrac{ca}{a+b}\right)$
$⇔P \leq \dfrac{1}{4}\left( \dfrac{b(a+c)}{a+c}+\dfrac{a(b+c)}{b+c}+\dfrac{c(a+b)}{a+b}\right)$
$⇔P \leq \dfrac{1}{4}\left( a+b+c\right)$ (đpcm)
