Let the inequality x ^ 4 + 6x ^ 2 + a > 4x ^ 3 + 8x hold for all real numbers X. try to determine the value range of A

Let the inequality x ^ 4 + 6x ^ 2 + a > 4x ^ 3 + 8x hold for all real numbers X. try to determine the value range of A

x^4+6x^2+a>4x^3+8x
＝〉x^4-4x³+6x²-8x>-a
＝〉x^4-2x³-2x³+4x²+2x²-4x-4x+8>8-a
＝〉x³(x-2)-2x²(x-2)+2x(x-2)-4(x-2)>8-a
＝〉(x³-2x²+2x-4)(x-2)>8-a
＝〉[x²(x-2)+2(x-2)](x-2)>8-a
＝〉(x²+2)(x-2)²>8-a
∵x²+2≥2,(x-2)²≥0
The minimum value of (X & sup2; + 2) (X-2) & sup2; is 0,
In order for the original inequality to hold, it must be 8-A, which is smaller than the minimum value of (X & sup2; + 2) (X-2) & sup2;, that is 8-a8

It is proved that the value of X & # 178; - 8x + 22 is always greater than 0 X & # 178; - 8x + 22 > 0 (x-4) > - 6
-

x²-8x+22
=x²-8x+16+6
=(x-4)²+6>0
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It is proved that the value of - 9x + 8x-2 is always less than 0

One qx2 + 8x-2 + 16-16 = 0
-9Χ2+8×+14-19=0

It is proved that the square of (1) - 9x + 8x-2 is always less than 0

=-9(x²-8x/9+16/81)+16/9-2
=-9(x-4/9)²-2/9
∵-9(x-4/9)²