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Algebra Difficulty 6.1 National Olympiad Find the answer

Given that a,b,c,d,ea,b,c,d,e are real numbers such that
a+b+c+d+e=8a+b+c+d+e=8 ,
a2+b2+c2+d2+e2=16a^2+b^2+c^2+d^2+e^2=16 .
Determine the maximum value of ee .

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

By Cauchy Schwarz, we can see that (1+1+1+1)(a2+b2+c2+d2)≥(a+b+c+d)2(1+1+1+1)(a^2+b^2+c^2+d^2)\geq (a+b+c+d)^2 thus 4(16−e2)≥(8−e)24(16-e^2)\geq (8-e)^2 Finally, e(5e−16)≥0e(5e-16) \geq 0 which means 165≥e≥0\frac{16}{5} \geq e \geq 0 so the maximum value of ee is 165\frac{16}{5} .
from: Image from Gon Mathcenter.net

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