K is a real number that satisfies the following property : "for every 3 positiv numbers ,a,b,c ; if a+b+c≤K then abc≤K " Can you find the biggest value of K ?

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Answer 1 · 2017-11-26T00:12:04

#K=3sqrt(3)#

If we put:

#a=b=c = K/3#

Then:

#abc = K^3/27 <= K#

So:

#K^2 <= 27#

So:

#K <= sqrt(27) = 3sqrt(3)#

If we have #a+b+c<=3sqrt(3)# then we can tell that the case #a=b=c=sqrt(3)# gives the maximum possible value of #abc#:

For example, if we fix #c in (0, 3sqrt(3))# and let #d = 3sqrt(3)-c#, then:

#a+b = d#

So:

#abc = a(d-a)c#

#color(white)(abc) = (ad-a^2)c#

#color(white)(abc) = (d^2/4-(a^2-2(a)(d/2)+(d/2)^2))c#

#color(white)(abc) = (d^2-(a-d/2)^2)c#

which has its maximum value when #a=d/2# and #b=d/2#, that is when #a=b#.

Similarly if we fix #b#, then we find the maximum is when #a=c#.

Hence the maximum value of #abc# is attained when #a=b=c#.

So #K=3sqrt(3)# is the maximum possible value of #a+b+c# such that #abc <= K#