=Đáp án: ta có :
p=(2a +3b+3c+1/2015+a)+(3a+2b+3c/2016+b)+(3a+2b+3c/2016+b)
p+3=(2a +3b+3c+1/2015+a)+1+(3a+2b+3c/2016+b)+1+(3a+2b+3c/2016+b)+1
=>(3a +3b+3c+2016/2015+a)+(3a+3b+3c+2016/2016+b)+(3a+3b+3c+2016/2016+b)
=>(3a+3b+3c+2016) ((1/15+a)(1/16+b)(1/17+c))
=>4*2016((1/15+a)(1/16+b)(1/17+c)) (a+b+c=2016)
8064=((1/15+a)(1/16+b)(1/17+c))
ví abc dương ta áp dụng bdt ta có ;
((1/15+a)(1/16+b)(1/17+c)) lớn hơn hc bằng 9/2015+2016+2017+a+b+c=9/8640
=>p+3 lớn hơn hc bằng 8064*9/8064= 9 =>p lớn hơn hc bằng 6
<=>2015+a=2016+b+2017+c
a+b+c=2016
<=>a=673;b=672;c=671
