giải
cho a,b,c là các số thực dương thoả mãn :$a+b+c=1$ chứng minh $\dfrac{ab}{c+1} +\dfrac{bc}{a+1} +\dfrac{ca}{b+1}≤\dfrac{1}{4}$
Giải:
$\dfrac{ab}{c+1}=\dfrac{ab}{(c+a)+(b+c)}=ab.\dfrac{1}{(c+a)+(b+c)}≤\dfrac{ab}{4}.(\dfrac{1}{c+a}+ \dfrac{1}{b+c})$
Tương tự, $\dfrac{bc}{a+1}≤\dfrac{bc}{4}.(\dfrac{1}{a+b}+ \dfrac{1}{a+c}), \dfrac{ca}{b+1}≤\dfrac{ca}{4}.(\dfrac{1}{a+b}+ \dfrac{1}{b+c})$
Suy ra $\dfrac{ab}{c+1}+\dfrac{bc}{a+1}+\dfrac{ca}{b+1} ≤ \dfrac{ab}{4}.(\dfrac{1}{c+a}+ \dfrac{1}{b+c})+\dfrac{bc}{4}.(\dfrac{1}{a+b}+ \dfrac{1}{a+c})+\dfrac{ca}{4}.(\dfrac{1}{a+b}+ \dfrac{1}{b+c})$
Lại có $\dfrac{ab}{4}.(\dfrac{1}{c+a}+ \dfrac{1}{b+c})+\dfrac{bc}{4}.(\dfrac{1}{a+b}+ \dfrac{1}{a+c})+\dfrac{ca}{4}.(\dfrac{1}{a+b}+ \dfrac{1}{b+c})=\dfrac{ac+cb}{4(a+b)}+\dfrac{ba+cb}{4(b+c)}+\dfrac{cb+ab}{4(c+a)}$
$=\dfrac{a}{4}+\dfrac{b}{4}+\dfrac{c}{4}=\dfrac{a+b+c}{4}=\dfrac{1}{4}$
$⇒$ ĐPCM
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