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Let $a_{i}>0,i=1,2,\cdots,n$,

If $$1+\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}\ge k\;\left(\dfrac{1}{\sqrt{1+a^2_{1}}}+\dfrac{1}{\sqrt{1+a^2_{1}+a^2_{2}}}+\cdots+\dfrac{1}{\sqrt{1+a^2_{1}+\cdots+a^2_{n}}}\right)$$

then find the maximum value for $k$

My idea was to use

$$\left(\sum_{i=1}^{n}\dfrac{1}{\sqrt{1+a^2_{1}+a^2_{2}+\cdots+a^2_{i}}}\right)^2\le\left[\sum_{i=1}^{n}\dfrac{a_{i}}{1+a^2_{1}+a^2_{2}+\cdots+a^2_{i}}\right]\left(\sum_{i=1}^{n}\dfrac{1}{a_{i}}\right)$$ and to use the approaches discussed here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=119163&sid=cd139cff36102a55e4f91a7c8dcd4f21#p119163

Alas! I failed to figure out how to do it.

Ahmed Masud
  • 490
math110
  • 93,304

2 Answers2

9

Substituting $a_i=\sqrt{2}^{i-1}$ gives

\begin{align} 1+\frac{1-\frac{1}{\sqrt{2}^n}}{1-\frac{1}{\sqrt{2}}}=1+\sum_{i=1}^{n}{(\frac{1}{\sqrt{2}})^{i-1}}=1+\sum_{i=1}^{n}{\frac{1}{a_i}} & \geq k\left(\sum_{i=1}^{n}{\frac{1}{\sqrt{1+\sum_{j=1}^{i}{a_j^2}}}}\right)\\ & =k\left(\sum_{i=1}^{n}{\frac{1}{\sqrt{1+\sum_{j=1}^{i}{2^{j-1}}}}}\right) \\ & =k\left(\sum_{i=1}^{n}{\frac{1}{\sqrt{2^i}}}\right) \\ & =k\frac{1}{\sqrt{2}}\frac{1-\frac{1}{\sqrt{2}^n}}{1-\frac{1}{\sqrt{2}}} \end{align}

Thus \begin{align} k \leq \frac{1+\frac{1-\frac{1}{\sqrt{2}^n}}{1-\frac{1}{\sqrt{2}}}}{\frac{1}{\sqrt{2}}\frac{1-\frac{1}{\sqrt{2}^n}}{1-\frac{1}{\sqrt{2}}}}=\sqrt{2}+\frac{1}{\frac{1}{\sqrt{2}}\frac{1-\frac{1}{\sqrt{2}^n}}{1-\frac{1}{\sqrt{2}}}}=\sqrt{2}+\frac{\sqrt{2}-1}{1-\frac{1}{\sqrt{2}^n}} \end{align}

Since this must hold for all $n$, we have $k \leq 2\sqrt{2}-1$. We now show that $k=2\sqrt{2}-1$ works, so that $2 \sqrt{2}-1$ is the maximum value of $k$. In other words, we will show that

$$1+\sum_{i=1}^{n}{\frac{1}{a_i}} \geq (2\sqrt{2}-1)\left(\sum_{i=1}^{n}{\frac{1}{\sqrt{1+\sum_{j=1}^{i}{a_j^2}}}}\right)$$

By Cauchy Schwarz inequality we have

$$\sqrt{2}^i\sqrt{1+\sum_{j=1}^{i}{a_j^2}}=\sqrt{1+\sum_{j=1}^{i}{a_j^2}}\sqrt{1+\sum_{j=1}^{i}{2^{j-1}}} \geq 1+\sum_{j=1}^{i}{\sqrt{2}^{j-1}a_j}$$

By weighted AM-HM inequality we have

$$\frac{1+\sum_{j=1}^{i}{\sqrt{2}^{j-1}a_j}}{2^i}=\frac{1+\sum_{j=1}^{i}{2^{j-1}\frac{a_j}{\sqrt{2}^{j-1}}}}{1+\sum_{j=1}^{i}{2^{j-1}}} \geq \frac{1+\sum_{j=1}^{i}{2^{j-1}}}{1+\sum_{j=1}^{i}{2^{j-1}\frac{\sqrt{2}^{j-1}}{a_j}}}=\frac{2^i}{1+\sum_{j=1}^{i}{\frac{(2\sqrt{2})^{j-1}}{a_j}}}$$

Thus

\begin{align} \sum_{i=1}^{n}{\frac{1}{\sqrt{1+\sum_{j=1}^{i}{a_j^2}}}} \leq \sum_{i=1}^{n}{\frac{\sqrt{2}^i}{1+\sum_{j=1}^{i}{\sqrt{2}^{j-1}a_j}}} & \leq \sum_{i=1}^{n}{\frac{1+\sum_{j=1}^{i}{\frac{(2\sqrt{2})^{j-1}}{a_j}}}{(2\sqrt{2})^i}} \\ &=\sum_{i=1}^{n}{\frac{1}{(2\sqrt{2})^i}}+\sum_{j=1}^{n}{\frac{(2\sqrt{2})^{j-1}}{a_j}\sum_{i=j}^{n}{\frac{1}{(2\sqrt{2})^i}}} \\ &=\sum_{i=1}^{n}{\frac{1}{(2\sqrt{2})^i}}+\sum_{j=1}^{n}{\frac{1}{a_j}\sum_{i=1}^{n-j+1}{\frac{1}{(2\sqrt{2})^i}}} \\ & \leq \sum_{i=1}^{\infty}{\frac{1}{(2\sqrt{2})^i}}+\sum_{j=1}^{n}{\frac{1}{a_j}\sum_{i=1}^{\infty}{\frac{1}{(2\sqrt{2})^i}}} \\ &=\frac{1}{2\sqrt{2}-1}(1+\sum_{i=1}^{n}{\frac{1}{a_i}}) \end{align}

Ivan Loh
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1

Let's reduce the statements:

$$(1):\qquad 1+\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n} \quad=\quad 1+\sum_{i = 1}^{n}\frac{1}{a_i}$$

$$(2):\qquad k\left( \frac{1}{\sqrt{1+{a_1}^2}}+...+\frac{1}{\sqrt{1+{a_1}^2+...+{a_n}^2}} \right) \quad=\quad k\left ( \sum_{i=1}^{n} \frac{1}{\sqrt{(1+\left( \sum_{j = 1}^{i}{{a_j}^2}\right )}}\right )$$

Now, with further reduction, you notice, as $(2)$ implies:

$$(3):\qquad k\left ( \sum_{i=1}^{n} \frac{1}{\sqrt{(1+\left( \sum_{j = 1}^{i}{{a_j}^2}\right )}}\right ) = \frac{k}{1-k}$$

and:

$$(4):\qquad 1+\sum_{i = 1}^{n}\frac{1}{a_i}\ge\frac{k}{1-k} \Rightarrow$$ $$(5):\qquad 1+\sum_{i = 1}^{n}\frac{1}{a_i} + \frac{k}{k-1}\ge0$$

Thus we have:

$$(6):\qquad 1+\sum_{i = 1}^{n}\frac{1}{a_i} \ge k\left ( \sum_{i=1}^{n} \frac{1}{\sqrt{(1+\left( \sum_{j = 1}^{i}{{a_j}^2}\right )}}\right ) \Rightarrow$$ $$1+\sum_{i = 1}^{n}\frac{1}{a_i} + \frac{k}{k-1}\ge0$$

JohnWO
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