Inequality (a+b)(a+b+c)&lt;0 Prove (b−c)2&gt;4a(a+b+c)

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Inequality  (a+b)(a+b+c)&amp;lt;0  Prove  (b−c)2&amp;gt;4a(a+b+c)

Rosa Nicholson · Accepted Answer

Step 1  Consider the quadratic  f(x)=ax2−(b−c)x+(a+b+c)  f(1)f(0)=2(a+c)(a+b+c)&amp;lt;0  Thus if a≠0, then this has a real root in (0,1) and so  (b−c)2≥4a(a+b+c)  If (b−c)2+4a(a+b+c),  then we have a double root in (0,1) in which case, f(0) and f(1) will have the same sign.  Thus  (b−c)2&amp;gt;4a(a+b+c)  If a=0, then c(b+c)&amp;lt;0, and so we cannot have b=c and thus (b−c)2&amp;gt;0=4a(a+b+c)  And if you want a more direct approach, we show that (p+q+r)r&amp;lt;0⇒q2&amp;gt;4pr using the following identity:  (p+q+r)r=(p(1+q2p)q2p)2+(r−q24p)2  +p(r−q24p)((1+q2p)2+(q2p)2)  If (p+q+r)r&amp;lt;0, then we must have have  p(r−q24p)&amp;lt;0  as all the other terms on the right side are non-negative.  Of course, this was gotten by completing the square in px2+qx+r and setting x=0 and x=1 and multiplying.

caoireoilns · Accepted Answer

Because  (b−c)2−4a(a+b+c)  =(b−c)2+(4a+8c)(a+b+c)−8(a+c)(a+b+c)  =(2a+b+3x)2−8(a+c)(a+b+c)&amp;gt;0

Inequality (a+b)(a+b+c)<0 Prove (b-c)^{2}>4a(a+b+c)

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