Nice use of AM GM

Calculus Level 5

We are given 4 lines with equations:

  • y=(2+3a2+3b2(1+a2)2+b(6a+6a3)a+b(1+a2))min2y = \Bigg( \dfrac{2 + 3a^2 + 3b^2(1 + a^2)^2 + b(6a + 6a^3)}{a + b(1 + a^2)} \Bigg)^2_{\min}
  • y=(c2d22c2d+2c2+2cd2c+1c2d+c)min+38 y = \Bigg(\dfrac{c^2d^2 - 2c^2d + 2c^2 + 2cd - 2c + 1}{c^2d + c} \Bigg)_{\min} + 3 - \sqrt{8}
  • y=[e]x2 y = [e]x^2
  • y=12[e]x2 y =\dfrac{1}{2} [e]x^2

With a,b,c,d>0,e1 a,b,c,d > 0, e\geq 1 .

The area bounded by the given lines and the curves when obtain a maximum value of MM, with M=a1a2(a31)(a4a51) M = \dfrac{a_{1}}{a_{2}}(\sqrt{a_{3}} - 1)(a_{4}\sqrt{a_{5}} - 1).

Find a1+a2+a3+a4+a5 a_{1} + a_{2} + a_{3} + a_{4} + a_{5}.

Details and Assumptions:

  • [.] [.] denote the greatest integer function.
  • (f(x))min (f(x))_{\text{min}} denote the minimum value of f(x)f(x) .
Dedicated to Aditya Raut and Guilherme Dela Corte.
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