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Prove that the inequality
x(x+ 1)(x+ 2)(x+ 3) ≥ −1
holds for all real numbers x.

Sagot :

pepe11
(x²+x)(x²+5x+6) = x^4 + 6x³ + 11x² + 6x  + 1 ≥0
f(x) = x^4 + 6x³ + 11x² + 6x  + 1
f'(x) = 4x³ + 18x² +22x +6 
racines x1 = -0,301 x2 = -1,5 x3 = -2,618
variations 

x                                  -2,62                    -1,5                   -0,3
f'(x)                      -          0             +          0           -            0          +
f(x)    -inf            \            0             /                        \             0           /

on voit que cette fonction st toujours positive
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