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vansang02121998

Áp dụng Cauchy

$\dfrac{ab}{a+c}+\dfrac{ab}{b+c} \ge \dfrac{4ab}{a+b+2c}$

$\dfrac{bc}{a+b}+\dfrac{bc}{a+c} \ge \dfrac{4bc}{2a+b+c}$

$\dfrac{ac}{a+b}+\dfrac{ac}{b+c} \ge \dfrac{4ac}{a+2b+c}$

Cộng vế với vế, ta có

$\dfrac{ab+bc}{a+c}+\dfrac{ac+bc}{a+b}+\dfrac{ab+ac}{b+c} \ge \dfrac{4ab}{a+b+2c}+\dfrac{4bc}{2a+b+c}+\dfrac{4ac}{a+2b+c}$

$\leftrightarrow \dfrac{a+b+c}{4} \ge \dfrac{ab}{a+b+2c}+\dfrac{ac}{a+2b+c}+\dfrac{bc}{2a+b+c}$
 
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