Part. a) $(a^2+b^2+c^2)(a^4+b^4+c^4) \ge (a^3+b^3+c^3)^2$ Sup. Yes $13*40.5 \ge 28^2$ $526.5 \ge784$ ¡Contradiction! Part. b) $a+b+c=6$ $ab+bc+ac=\frac{23}{2}$ $abc=7$ $2x^3 - 12x^2 +23x -14=0$ Solutions: $(a,b,c)=(2,2+\frac{1}{\sqrt2},2-\frac{1}{\sqrt2})$ It's clear. Answer: Yes

Mathsy123 wrote: Part. a) $(a^2+b^2+c^2)(a^4+b^4+c^4) \ge (a^3+b^3+c^3)^2$ Sup. Yes $13*40.5 \ge 28^2$ $526.5 \ge784$ ¡Contradiction! Part. b) $a+b+c=6$ $ab+bc+ac=\frac{23}{2}$ $abc=7$ $2x^3 - 12x^2 +23x -14=0$ Solutions: $(a,b,c)=(2,2+\frac{1}{\sqrt2},2-\frac{1}{\sqrt2})$ It's clear. Answer: Yes A little bit easier for part a) From Cauchy we would have $(a+b+c)(a^3+b^3+c^3)\geq (a^2+b^2+c^2)^2\Rightarrow 6\cdot 28\geq 13^2\Rightarrow 168\geq 169$ A contradiction.

Mathsy123 wrote: $2x^3 - 12x^2 +23x -14=0$ I don't get how we got to this part I know that this is probably used with uvw method and I don't know it so could you give me a bit more elaborate explanation. Thanks

White-Wolf wrote: Mathsy123 wrote: $2x^3 - 12x^2 +23x -14=0$ I don't get this part I know that this is probably used with uvw method and I don't know it so could you give me a bit more elaborate explanation. Thanks Well known identities and Vieta's formulas (constructing a polynomial with zeros a, b and c).

I asked because I don't know it,that is what I read everywhere I just want an explanation to the problem.

Part (a) solutions are the six permutations of the numbers: $2-\frac{1}{6}\left(\sqrt[3]{72+6 \sqrt{138}}+\sqrt[3]{72-6 \sqrt{138}}\right)$ and $2+\frac{1}{12}\left(\sqrt[3]{72+6 \sqrt{138}}+\sqrt[3]{72-6 \sqrt{138}}\right)\pm\frac{\sqrt{3}}{12}\left(\sqrt[3]{72+6 \sqrt{138}}-\sqrt[3]{72-6 \sqrt{138}}\right)i$ Part (b) solutions are the six permutations of the numbers $2$ and $2\pm\frac{\sqrt{2}}{2}$ Therefore (a)=NO and (b)=YES.

Thanks a lot but I think you don't understand the question I don't know how from: $a+b+c=6$ $ab+bc+ac=\frac{23}{2}$ $abc$=7 to $2x^3 - 12x^2 +23x -14=0$ I wanna know the process.

$P(x)=2(x-a)(x-b)(x-c)$ - polynomial with roots $a,b, c$ Using Vieta's formulas for third degree polynomial we get: $P(x)=2x^3 - 12x^2 +23x -14$ Now we can see that one of its zeros is 2. After this, it is not hard to find remaining zeros.

Thank you a lot know I understand it.