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Prove that for $n\ge 2$, $n\in \mathbb{N}$, $a_{i}, b_{i}\ge 0$


When $n=2$, I can easily show that


I want to use Induction to prove this. Any suggestions on how to solve it?

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I don't think you should use induction on this. – Euclidean Mar 25 '13 at 7:20
up vote 2 down vote accepted

Fix $a_2,\cdots,a_n, b_2,\cdots, b_n$. Consider $b_1' = t$, $a_1'^2 = a_1^2 + b_1^2 - t^2$, for $t \in [0,\sqrt{a_1^2+b_1^2}]$. With this pair $(a_1',b_1')$ replacing $(a_1,b_1)$, one sees that right hand side is unchanged, and left hand side is a concave quadratic function in $t$. This means that LHS attains its minimum when $t$ hits the boundary, i.e. $(a_1,b_1)$ being replaced by $(\sqrt{a_1^2+b_1^2}, 0)$, or $(0, \sqrt{a_1^2+b_1^2})$.

This reduces the problem to the following one. Let $x_i^2 = a_i^2 + b_i^2$, $x_i \ge 0$. For each index $i$, $(a_i,b_i) = (x_i,0)$ or $(0,x_i)$. Rearrange the indices so that the first $k$ indices satisfy $(a_i,b_i) = (x_i,0)$, and the last $(n-k)$ satisfies $(a_i,b_i) = (0,x_i)$. We then need to show


for any $x_i \ge 0$.

Apply AM-GM for the two terms on the left separately, we get LHS is at least

$$\frac{k}{n} \left(\dfrac{n}{n-1}\right)^{n-1} (x_1\cdots x_k)^{1/k} + \frac{(n-k)^2}{n^2} (x_{k+1}\cdots x_n)^{1/(n-k)}$$

The inequality is clearly true when $k = 0$ or $n$. Otherwise apply AMGM again, with suitable weights, we get that this is at least $$\left(\dfrac{n}{n-1}\right)^{k(n-1)/n}\left(\frac{n-k}{n}\right)^{(n-k)/n}(x_1\cdots x_n)^{2/n}$$ So it suffices to show that $$\left(\dfrac{n}{n-1}\right)^{n-1}\left(\frac{n-k}{n}\right)^{(n-k)/k} \ge 1$$ Equivalently, $$\left(\dfrac{n}{n-1}\right)^{n-1} \ge \left(\frac{n}{n-k}\right)^{(n-k)/k} = \left(\frac{1}{1 - k/n}\right)^{n/k - 1}$$ Hence consider the function $f(u) = (1-u)^{u-1}$ for $u \in (0,1)$. It suffices to show that $f(u)$ is increasing in this range, which is true from the plot at wolframalpha.

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Oh,very nice, your methods Now I can't find any mistake, Thank you very much,and other have any idea? – math110 Mar 25 '13 at 10:02

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