Inequalities

**xwrathbringerx** · July 30th 2008, 11:31 PM

a) Prove that the inequality x(x+1)(x+2)(x+3) >= -1 holds for all real numbers x.

b) Prove that (ab+bc+ca)^2 >= 3abc(a+b+c) holds for all real numbers a, b and c

**NonCommAlg** · July 31st 2008, 03:56 AM

Originally Posted by xwrathbringerx

a) Prove that the inequality x(x+1)(x+2)(x+3) >= -1 holds for all real numbers x.

obviously $\left(t^2 - \frac{5}{4} \right)^2 \geq 0, \ \forall t \in \mathbb{R},$ which can be rewritten as: $\left(t^2 - \frac{9}{4} \right) \left(t^2 - \frac{1}{4} \right) \geq -1.$

thus: $\left(t-\frac{3}{2} \right) \left(t + \frac{3}{2} \right) \left(t - \frac{1}{2}\right) \left(t + \frac{1}{2} \right) \geq -1, \ \forall t \in \mathbb{R}.$ now put $t=x+\frac{3}{2}. \ \ \ \square$

b) Prove that (ab+bc+ca)^2 >= 3abc(a+b+c) holds for all real numbers a, b and c

Fact: for any real numbers $x,y,z: \ (x+y+z)^2 \geq 3(xy+yz+zx).$

Proof: $(x+y+z)^2-3(xy+yz+zx)=\frac{1}{2}((x-y)^2+(y-z)^2+(z-x)^2) \geq 0. \ \ \ \square$

now your inequality follows quickly if you put $x=ab, \ y=bc, \ z=ca.$

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