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$a,b,c,x_1,x_2,x_3,...,x_n>0, a+b+c=1,\displaystyle \prod_{i=1}^n x_i=1 $ . Prove that $$(ax_1^2+bx_1+c)...(ax_n^2+bx_n+c)\geq1$$.

I've tried just writing out as a product using the product sign and then to group certain parts of the product. Nothing came out puf it what we need. Also tried induction with the base $n=1,2$ rather straightforward. How is it done?

Suzu Hirose
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Jackie Poehler
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  • If $a = b = c = \frac{1}{4}$ and $x_1 = x_2 = \cdots = x_n = 1$, then the left side is $(3/4)^n$ which is less than $1$. Are you sure you aren't missing any conditions on $a,b,c$? – JimmyK4542 Dec 26 '14 at 08:18
  • are you sure this is right? let $a=b=c=x_{i}=\dfrac{1}{1000}$ – math110 Dec 26 '14 at 08:20

2 Answers2

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Notice $a, b, c > 0$ and $a + b + c = 1$. By AM $\ge$ GM, we have

$$a x_i^2 + b x_i + c \ge (x_i^2)^a (x_i)^b 1^c = x_i^{2a+b} \quad\text{ for each }\;i = 1, \ldots, n $$ As a result,

$$\prod_{i=1}^n (a x_i^2 + b x_i + c) \ge \prod_{i=1}^n x_i^{2a+b} = \left(\prod_{i=1}^n x_i \right)^{2a+b} = 1^{2a+b} = 1$$

achille hui
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if $$\prod_{i=1}^{n}x_{i}=1,a+b+c=1$$

then Use Hölder's inequality $$(ax^2_{1}+bx_{1}+c)(ax^2_{2}+bx_{2}+c)\cdots (ax^2_{n}+bx_{n}+c)\ge \\(a(x_{1}x_{2}\cdots x_{n})^{\frac{2}{n}}+b(x_{1}x_{2}\cdots x_{n})^{\frac{1}{n}}+c)^n=(a+b+c)^n=1$$

r9m
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math110
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