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So is high school math.
The answer given by Sun Zhenwei 1 14 cannot be established.

Where a+b ≧ 2 √ (AB) must be based on the premise that both A and B are non-negative, but it is not in the condition and cannot be deduced.

Here is a reasonable solution:

∵a+b+c= 1、a^2+b^2+c^2= 1,∴a+b= 1-c、a^2+b^2= 1-c^2。

Introduce the function: f (x) = (x+a) 2+(x+b) 2.

∵a>b,∴f(x)>0。

And f (x) = (x2+2ax+a2)+(x2+2bx+B2) = 2x2+2 (a+b) x+(a2+B2),

∴f(x)=2x^2+2( 1-c)x+( 1-c^2)。

Obviously, f(x) is a parabola with an upward opening, and f (x) > 0.

The discriminant of equation 2x 2+2 (1-c) x+(1-c 2) = 0 < 0, ∴ 4 (1-c) 2-8 (1-c).

∴( 1-c)^2-2( 1-c^2)