Cho $a,b,c,d,e$ là các số thực dương có tổng bằng 5. Chứng minh rằng $abc+bcd+cde+dea+eab\leq 5$
#1
Posted 11-06-2018 - 14:51
#2
Posted 11-06-2018 - 15:14
$$\underbrace{\left ( a+ b+ c+ d+ e \right )\left ( b+ c+ d+ e+ a \right )\left ( c+ d+ e+ a+ b \right )\geqq 5\,5\,\left ( abc+ bcd+ cde+ dea+ eab \right ) }_{Holder}$$
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